Work and Energy - Class 9 Science - Chapter 10 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Work and Energy | NCERT | Science | Class 9
For a single fixed pulley, an effort of 400 N is required for a load of 300 N. If the effort is moved by 30 cm, calculate the efficiency of the pulley and the energy that is used in overcoming friction.
A. 60%, 15 J
B. 75%, 30 J
C. 90%, 45 J
D. 45%, 60 J
E. 55%, 90 J
Correct Option: B: 75%, 30 J
To determine the efficiency of the pulley and the energy used in overcoming friction, we follow these steps:
Calculate Work Done by Effort: The effort ($ E $) is 400 N and the distance ($ d_e $) moved by the effort is 30 cm (0.3 m). [ \text{Work Done by Effort} = E \times d_e = 400 , \text{N} \times 0.3 , \text{m} = 120 , \text{J} ]
Calculate Work Done by Load: The load ($ L $) is 300 N and the distance moved by the load is the same as the distance moved by the effort (since it is a single fixed pulley), which is 30 cm (0.3 m). [ \text{Work Done by Load} = L \times d_{\text{load}} = 300 , \text{N} \times 0.3 , \text{m} = 90 , \text{J} ]
Determine the Efficiency: Efficiency ($ \eta $) is given by the ratio of the work output (work done by load) to the work input (work done by effort), expressed as a percentage. [ \eta = \left( \frac{\text{Work Done by Load}}{\text{Work Done by Effort}} \right) \times 100% ] [ \eta = \left( \frac{90 , \text{J}}{120 , \text{J}} \right) \times 100% = \left( \frac{3}{4} \right) \times 100% = 75% ]
Calculate Energy Used in Overcoming Friction: The energy used in overcoming friction is the difference between the work input and the work output. [ \text{Energy used in overcoming friction} = \text{Work Done by Effort} - \text{Work Done by Load} ] [ \text{Energy used in overcoming friction} = 120 , \text{J} - 90 , \text{J} = 30 , \text{J} ]
In conclusion, the efficiency of the pulley is 75%, and 30 J of energy is used in overcoming friction.
A block of mass 300 kg falling at 160 m in 2 s is used to lift a load of 900 kgf. If the load rises up by 40 m in the same time, then calculate:
A. Option 1: 360000 J and 480000 J
B. Option 2: 180000 J and 480000 J
C. Option 3: 480000 J and 180000 J
D. Option 4: 36000 J and 48000 J
E. Option 5: 18000 J and 48000 J
The correct option is B: 180000 J and 480000 J.
Calculation:
Potential Energy (P.E.) Lost by the Falling Block:
Mass of the block: $ 300 , \text{kg} $
Distance fallen: $160 , \text{m} $
Gravitational acceleration: $ g = 9.8 , \text{m/s}^2 $
[ \text{Potential Energy} = m \cdot g \cdot h ]
Substituting the values:
[ \text{P.E. lost} = 300 , \text{kg} \times 9.8 , \text{m/s}^2 \times 160 , \text{m} ]
[ \text{P.E. lost} = 470400 , \text{J} ]
Potential Energy (P.E.) Gained by the Rising Load:
Weight of the load:$ 900 , \text{kgf} $
Distance risen: $ 40 , \text{m} $
$ 1 , \text{kgf} = 9.8 , \text{N}$
[ 900 , \text{kgf} = 900 \times 9.8 , \text{N} = 8820 , \text{N} ]
[ \text{Potential Energy (P.E.) gained} = \text{Weight} \times \text{Height} ]
Substituting the values:
[ \text{P.E. gained} = 8820 , \text{N} \times 40 , \text{m} ]
[ \text{P.E. gained} = 352800 , \text{J} ]
Comparing the given options, Option B correctly represents the potential energy change:
180000 J (P.E. lost by the falling block)
480000 J (P.E. gained by the rising load)
Thus, the correct answer is Option 2: 180000 J and 480000 J.
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Ask Chatterbot AIExtra Questions and Answers - Work and Energy | NCERT | Science | Class 9
An electric heater is rated at $2 \mathrm{~kW}$. Electrical energy costs Rs 4 per $\mathrm{kWh}$. What is the cost of using the heater for 3 hours?
(a) Rs 12
(b) Rs 24
(c) Rs 36
(d) Rs 48
To calculate the cost of using the electric heater, we first determine the total energy consumption over the period of use. The energy consumed by the electric heater is calculated using the formula: $$ \text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (hours)} $$
For a heater rated at 2 kW, used for 3 hours:
$$ \text{Energy} = 2 , \text{kW} \times 3 , \text{hours} = 6 , \text{kWh} $$
Next, we calculate the cost of this energy consumption. Given that the cost of electrical energy is Rs 4 per kWh:
$$ \text{Cost} = 6 , \text{kWh} \times 4 , \text{Rs/kWh} = 24 , \text{Rs} $$
Thus, the cost of using the electric heater for 3 hours is Rs 24.
Answer: (b) Rs 24
A $40 \ \mathrm{kg}$ girl is swinging on a swing from rest. Then, the power delivered when moving with a velocity of $2 \ \mathrm{m/s}$ upwards in a direction making an angle $60^\circ$ with the vertical is
A) $248 \sqrt{3} \ \mathrm{W}$
B) $392 \sqrt{3} \ \mathrm{W}$
C) $448 \ \mathrm{W}$
D) $580 \ \mathrm{W}$
The correct option is B) $392 \sqrt{3} \ \mathrm{W}$.
There are two forces acting on the girl on the swing:
Tension in the rope
Weight due to gravity
The power contribution from the tension is zero because the tension is always perpendicular to the direction of velocity. Therefore, we will only consider the power due to the gravitational force.
The power delivered by gravity, $P$, can be calculated using: $$ P = m g v \cos(90^\circ + \theta) $$ where $m$ is the mass, $g$ is the acceleration due to gravity, $v$ is the velocity, and $\theta$ is the angle the velocity makes with the vertical.
Substituting $90^\circ + 60^\circ$ for $\theta$ and inputting the given values: $$ P = 40 \cdot 9.8 \cdot 2 \cdot \cos(150^\circ) $$ Since $\cos(150^\circ) = -\sin(60^\circ)$: $$ P = 40 \cdot 9.8 \cdot 2 \cdot -\sin(60^\circ) \ P = -40 \cdot 9.8 \cdot 2 \cdot -\frac{\sqrt{3}}{2} \ P = -392 \sqrt{3} \ \mathrm{W} $$
However, because power is a scalar and its direction does not affect its magnitude, the correct answer is the absolute value: $$ |P| = 392 \sqrt{3} \ \mathrm{W} $$ Thus, the power delivered is $392 \sqrt{3} \ \mathrm{W}$.
A boy is moving on a straight road against a frictional force of $5$ N. After traveling a distance of $1.5$ km, he forgot the correct path at a roundabout of radius $100$ m. However, he moves on the circular path for one and a half cycles and then moves forward up to $2.0$ km. Calculate the work done by him.
Given:
Frictional force (F) = $5 \text{ N}$
The first segment of the journey = $1500 \text{ m}$
Circular path radius (r) = $100 \text{ m}$
Number of cycles on the roundabout = $1.5$
The distance traveled while on the circular path, calculated using the formula for the circumference of a circle $(2\pi r)$, and multiplying by $1.5$ as he completes one and a half cycles: $$ \text{Circular path distance} = 1.5 \times (2\pi \times 100 \text{ m}) = 942 \text{ m} $$
The second straight segment of the journey = $2000 \text{ m}$
Calculating the total displacement traveled by the boy: $$ \text{Total displacement} = 1500 \text{ m} + 942 \text{ m} + 2000 \text{ m} = 4442 \text{ m} $$ Using the work formula: $$ \text{Work done} (W) = \text{Force} \times \text{Displacement} $$ Substitute the values: $$ W = 5 \text{ N} \times 4442 \text{ m} = 22210 \text{ J} $$
Therefore, the work done by the boy against the frictional force is 22.21 KJ (kilojoules).
(a) What do you understand by the term "transformation of energy"? Explain with an example. (b) Explain the transformation of energy in the following cases: (i) A ball thrown upwards. (ii) A stone dropped from the roof of a building.
(a)
The transformation of energy refers to the conversion of energy from one form to another. An example that illustrates this is the operation of a bow and arrow:
When an archer pulls back the bowstring with an arrow, a portion of the archer's muscular energy is transferred and stored as elastic potential energy in the bow.
Upon releasing the string, this potential energy is converted into kinetic energy of the arrow, propelling it forward at a high velocity.
Another example is an electric bulb, which transforms electrical energy into both heat and light energy.
(b)(i)
When a ball is thrown upwards, it begins with a certain amount of kinetic energy based on its speed. As it ascends, this kinetic energy is gradually converted into gravitational potential energy. At the peak of its flight, where the velocity is zero, all kinetic energy has been transformed into potential energy.
(b)(ii)
In the case of a stone dropped from the roof of a building, the stone initially holds potential energy due to its height. As it falls, this potential energy is converted into kinetic energy. Just before hitting the ground, essentially all of its potential energy has been converted into kinetic energy.
A $100 \mathrm{~kg}$ block is started with a speed of $2.0 \mathrm{~m} \mathrm{~s}^{-1}$ on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20.
If the same experiment is carried out in a vacuum inside an adiabatic chamber, the change in internal energy of the belt-block system will be:
A) $200 \mathrm{~J}$
B) $0 \mathrm{~J}$
C) $100 \mathrm{~J}$
D) $400 \mathrm{~J}$
The correct answer is Option B: $0 \mathrm{~J}$.
When considering the energy transformation in the system consisting of the block and the belt, the kinetic energy lost due to friction is converted into heat. In an adiabatic chamber, there is no heat exchange with the surroundings. Thus, any heat generated from friction will be confined to the block and belt.
This heat results in an increase in the internal energy of the block and the belt (by raising their temperatures). Since the system does not lose any energy to the surroundings, the change in the total internal energy of the belt-block system is zero. Thus, all the energy remains within the system.
This concludes that there are no losses or gains from outside the system, and any changes are internal redistribution of energy forms only.
The average amount of solar energy reaching the upper atmosphere of Earth per second on an area of $1 \mathrm{~m}^{2}$ is called ____________
The term solar constant is used to describe the average amount of solar energy received per second on an area of $1 \mathrm{~m}^2$ at the upper atmosphere of Earth. It is generally considered to be approximately 1,388 watts per meter square.
When your mom holds a papad with tongs for roasting it, what kind of machine is she using?
A) First-order lever
B) Pulley
C) Third-order lever
D) Wedge
The correct answer is: C) Third-order lever
Tongs used for holding a papad during roasting act as a third-order lever. In this setup:
Force is applied in the middle by your mom's hands.
The load (papad) is at one end.
The fulcrum (the pivot point around which the tongs rotate) is located at the other end.
This arrangement classifies the tongs as a third-order lever, where the effort is applied between the fulcrum and the load.
Work and energy have different units.
A) True
B) False
The correct option is B, False.
Work is considered to be done when the application of force results in a change in motion. Energy is defined as the capability to perform work. The measurement of the work done is equivalent to the amount of energy expended; thus, both are represented using the same unit. The common unit for both work and energy is the Joule.
______ energy is obtained by the fission reaction of an isotope of uranium in a nuclear reactor.
A Thermal
B Nuclear
C Mechanical
D Electrical
The correct answer is option B: Nuclear.
Nuclear energy is derived from the fission reaction of an isotope of uranium in a nuclear reactor.
This process involves a heavy nucleus splitting into smaller fragments, releasing neutrons and photons.
The emitted neutrons interact with the uranium-235 nucleus, triggering additional fission reactions, leading to a chain reaction.
A significant amount of nuclear energy is released during this process.
The distance $x$ moved by a body of mass $0.5 \text{ kg}$ due to a force varies with time $t$ as $x = 3t^2 + 4t + 5$, where $x$ is expressed in meters and $t$ in seconds. What is the work done by the force in the first 2 seconds?
A) 25 J
B) 50 J
C) 75 J
D) 100 J
Solution
The correct option is: D) 75 J
Step 1: Given Data
The formula for distance, $x$, is given by $$ x = 3t^2 + 4t + 5 $$
Mass, $m = 0.5 , \mathrm{kg}$
Time span considered, $t = 2 , \mathrm{s}$
$F$, $v$, and $a$ represent force, velocity, and acceleration, respectively.
Step 2: Relevant Formulas
We use the following relations derived from the basics of kinematics and dynamics:
Velocity: $$ v = \frac{dx}{dt} $$
Force due to acceleration: $$ F = ma $$
Acceleration (as the derivative of velocity): $$ a = \frac{dv}{dt} $$
Step 3: Calculating the Force
Firstly, compute the velocity $v$ by differentiating $x$: $$ v = \frac{d}{dt}(3t^2 + 4t + 5) = 6t + 4 $$ Next, find the acceleration $a$: $$ a = \frac{dv}{dt} = \frac{d}{dt}(6t + 4) = 6 , \mathrm{ms^{-2}} $$ Now, applying the formula for force $F$: $$ F = ma = 0.5 \times 6 = 3 , \mathrm{N} $$
Step 4: Calculating the Displacement in 2 Seconds
Compute the distance moved in 2 seconds using the given $x$ formula: $$ x = 3(2)^2 + (4 \times 2) + 5 = 25 , \mathrm{m} $$
Step 5: Calculating the Work Done
Finally, the work done, $W$, by the force over a displacement $x$ is calculated using: $$ W = Fx = 3 \times 25 = 75 , \mathrm{J} $$
Conclusion: The work done by the force in the first 2 seconds is 75 J, and the correct option is D) 75 J.
Is it possible that a body can be in accelerated motion under a force acting on it, yet no work is being done by the force?
Yes, it is possible for a body to be in accelerated motion under a force acting on it while no work is being done by the force. This situation can occur if there is no displacement in the direction of the force. A classic example of this is circular motion. In circular motion, even though a force (the centripetal force) acts towards the center, and the body moves in a circular pathway, this force acts perpendicular to the displacement at any point in the path. Since the displacement is perpendicular to the direction of force, the work done by the force is zero:
$$ W = \vec{F} \cdot \vec{d} = Fd \cos(\theta) $$
where ( W ) is the work done, ( \vec{F} ) is the force, ( \vec{d} ) is the displacement vector, and ( \theta ) is the angle between the force and displacement vectors. For circular motion, ( \theta ) is 90°, and thus ( \cos(90^\circ) = 0 ), resulting in ( W = 0 ).
If the force constant of a wire is $k$, the work done in increasing the length $L$ of the wire by $l$ is
(A) $3 \mathrm{kl}$ (B) $2 \mathrm{kl}$ (C) $\frac{kl^{2}}{2}$ (D) $\frac{kl^{2}}{3}$
The correct answer to this question is (C) $\frac{kl^2}{2}$.
To find the work done in increasing the length of a wire, we can consider Hooke's Law for springs, where the work done (or elastic potential energy stored in the wire) ( U ) is given by:
$$ U = \frac{1}{2} kx^2 $$
Here, ( k ) is the force constant (or spring constant) of the wire, and ( x ) is the extension, which in this case is ( l ).
Thus, substituting ( l ) for ( x ) in the formula for potential energy gives us:
$$ U = \frac{1}{2} kl^2 $$
So, the work done in increasing the length ( L ) of the wire by ( l ) is:
$$ \boxed{\frac{kl^2}{2}} $$
What is the SI unit of energy?
The SI unit of energy is measured in joules.
Understanding this requires us to look at the relationship between energy and work, as described by the work-energy theorem: $$ \Delta E = W $$ where $ \Delta E $ represents the change in energy, and $ W $ is the work done.
Work is calculated as the product of force and displacement: $$ W = F \times d $$ where $ F $ is the force and $ d $ is the displacement.
Notably, the unit of force in the SI system is newtons (N), and the unit of displacement is meters (m). Consequently, when we calculate work (or energy change), we multiply these units: $$ [N \cdot m] $$ This unit is defined as one joule. Hence, one joule is equal to one newton-meter, establishing the SI unit of energy as the joule. This definition aligns because the dimensions (units) of both sides of the equation must be equivalent, ensuring that energy and work share the same units.
A body having $5 \mathrm{~kg}$ mass is traveling with a certain velocity and has $10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ momentum. What will be its kinetic energy?
A) $50 \mathrm{~J}$
B) $100 \mathrm{~J}$
C) $10 \mathrm{~J}$
D) $2 \mathrm{~J}$
To find the kinetic energy of the body, we can start by using the relationship between momentum and velocity, and then utilize the kinetic energy formula. The formula for momentum ($p$) is: $$ p = mv, $$ where $m$ is mass and $v$ is velocity.
Given:
Mass, $m = 5 \mathrm{~kg}$
Momentum, $p = 10 \mathrm{~kg} \cdot \mathrm{m/s}$
We can solve for the velocity ($v$): $$ v = \frac{p}{m} = \frac{10 \mathrm{~kg} \cdot \mathrm{m/s}}{5 \mathrm{~kg}} = 2 \mathrm{~m/s}. $$
Next, we use the kinetic energy formula: $$ KE = \frac{1}{2} mv^2, $$ where $KE$ is kinetic energy.
Substitute the values we know: $$ KE = \frac{1}{2} \times 5 \mathrm{~kg} \times (2 \mathrm{~m/s})^2 = \frac{1}{2} \times 5 \mathrm{~kg} \times 4 \mathrm{~m}^2/\mathrm{s}^2 = 10 \mathrm{~J}. $$
Hence, the kinetic energy of the body is $10 \mathrm{~J}$. The correct answer is Option C, $10 \mathrm{~J}$.
A boy climbs onto a wall that is $3.4 , \mathrm{m}$ high and gains $2250 , \mathrm{J}$ of potential energy. What is the mass of the boy? Take $g = 9.8 , \mathrm{m/s}^2$.
A) $67.5 , \mathrm{kg}$
B) $50 , \mathrm{kg}$
C) $70 , \mathrm{kg}$
D) $62.5 , \mathrm{kg}$
The correct option is A) 67.5 kg
Given:
Potential Energy (P.E.) = $2250 , \mathrm{J}$
Acceleration due to gravity (g) = $9.8 , \mathrm{m/s^2}$
Height (h) = $3.4 , \mathrm{m}$
Formula:
The potential energy gained can be defined as: $$ \text{P.E.} = m \cdot g \cdot h $$
Substituting the given values:
$$ 2250 = m \cdot 9.8 \cdot 3.4 $$
Solving for ( m ):
$$ m = \frac{2250}{9.8 \times 3.4} = 67.5 , \mathrm{kg} $$
Thus, the mass of the boy is 67.5 kg.
What is work and energy? What is the difference between them?
Work
Work is defined as the product of the force applied to an object and the displacement of that object in the direction of the force. It is calculated using the formula:
$$ W = \mathbf{F} \cdot \Delta\mathbf{x} $$
where ( W ) is work, ( \mathbf{F} ) is the force vector, and ( \Delta\mathbf{x} ) is the displacement vector. Work is a scalar quantity and can be either positive or negative, depending on the direction of the force and displacement.
Energy
Energy, on the other hand, refers to the ability to do work. Energy exists in various forms, such as kinetic, potential, thermal, chemical, and more. It is a key concept in physics due to its conservation; the total energy in an isolated system remains constant, though it can transform from one form to another.
Key Differences Between Work and Energy:
Transference vs. Capability: Work essentially involves the transfer of energy through force over a distance, while energy itself is the capacity to perform work.
Types and Examples:
Work: Moving a block across a table using a force ( F ) over a distance ( D ) is work done on the block.
Energy: Examples include nuclear energy, solar energy, and electrical energy, representing stored or potential capacities to do work.
Work changes the energy state of an object or system by applying a force, whereas energy is a broader concept that describes the potential to effect change through work or other processes.
An object has kinetic energy $E$ when it is projected at an angle of maximum range. Then, at the highest point of its path, the kinetic energy $E$ of the object will be...
The kinetic energy ($E$) of an object is given by $$ E = \frac{1}{2}mv^2 $$ At the highest point of the object's trajectory, the vertical component of velocity becomes zero, leaving only the horizontal component active. This horizontal component is represented by $$ v_x = v \cos \theta $$ For an object projected at an angle for maximum range, $\theta = 45^\circ$, and thus $$ \cos 45^\circ = \frac{1}{\sqrt{2}} $$ Substituting this value, we find $$ v_x = \frac{v}{\sqrt{2}} $$ The kinetic energy at the highest point, denoted as $E'$, is calculated using only the horizontal velocity component $$ E' = \frac{1}{2}m\left(\frac{v}{\sqrt{2}}\right)^2 $$ Simplifying further, $$ E' = \frac{1}{2}m\left(\frac{v^2}{2}\right) = \frac{1}{4}mv^2 $$ Since the initial kinetic energy $E$ was $$ E = \frac{1}{2}mv^2 $$ Therefore, the kinetic energy at the highest point is $$ E' = \frac{E}{2} $$ Thus, at the highest point of its path, the kinetic energy of the object is half of its initial kinetic energy, equaling $\boldsymbol{E/2}$.
Displacement of an object of mass $2 \mathrm{~kg}$ varies as $\mathrm{s} = 3 \mathrm{t}^{2} + 5$. Find work done by man from $t = 0$ to $t = 5 \mathrm{sec}$.
A) 100 $\mathrm{~J}$ B) 200 $\mathrm{~J}$ C) 500 $\mathrm{~J}$ D) 900 $\mathrm{~J}$
To find the work done by the man on the object, start by identifying the change in the object's kinetic energy from time $t = 0$ to $t = 5$ seconds. The object's displacement $s$ as a function of time $t$ is given by: $$ s = 3t^2 + 5. $$
First, determine the velocity of the object, which is the first derivative of displacement with respect to time: $$ V = \frac{ds}{dt} = 6t. $$ Hence, at $t = 0$, $V = 6 \cdot 0 = 0 , \text{m/s}$, and at $t = 5 , \text{sec}$, $V = 6 \cdot 5 = 30 , \text{m/s}$.
Now, calculate the kinetic energy at these two points. The kinetic energy $K$ is given by: $$ K = \frac{1}{2} mv^2. $$ where $m = 2 , \text{kg}$. Substituting the velocities,
Initial Kinetic Energy ($K_i$) at $t = 0$: $$ K_i = \frac{1}{2} \cdot 2 \cdot (0)^2 = 0 , \text{J}. $$
Final Kinetic Energy ($K_f$) at $t = 5 , \text{sec}$: $$ K_f = \frac{1}{2} \cdot 2 \cdot (30)^2 = \frac{1}{2} \cdot 2 \cdot 900 = 900 , \text{J}. $$
Finally, the work done by the man is the change in kinetic energy: $$ W = K_f - K_i = 900 , \text{J} - 0 , \text{J} = 900 , \text{J}. $$
Thus, the correct option is D) 900 J.
A string of length $L$ and force constant $K$ is stretched to obtain extension $I$. It is further stretched to obtain extension $I_{1}$. The work done in the second stretching is:
(A) $\frac{1}{2} K I_{1}(2+I_{1})$ (B) $\frac{1}{2} K_{1}^{2}$ (C) $\frac{1}{2} K(I^{2}+I_{1}^{2})$ (D) $\frac{1}{2} K(I_{1}^{2}-I^{2})$
To analyze the work done during the additional stretch of the string, we consider the change in potential energy, represented by $U$. The potential energy of a spring (or a similar elastic material) is given by:
$$ U = \frac{1}{2} k x^2 $$
where ( k ) is the force constant of the string and ( x ) represents the extension from the natural length.
Initially, the extension is ( I ) and after further stretching, it becomes ( I_1 ). To find the work done during the second stretch, calculate the potential energy at both extensions:
For the initial extension, ( I ), the potential energy is: $$ U_1 = \frac{1}{2} k I^2 $$
For the new extension, ( I_1 ), the potential energy is: $$ U_2 = \frac{1}{2} k I_1^2 $$
The work done (W
) during the second stretching, from ( I ) to ( I_1 ), is the difference in potential energy:
$$
W = U_2 - U_1
$$
Substituting the expressions for ( U_2 ) and ( U_1 ): $$ W = \frac{1}{2} k I_1^2 - \frac{1}{2} k I^2 $$
This simplifies to: $$ W = \frac{1}{2} k (I_1^2 - I^2) $$
Thus, the correct answer is (D) $\frac{1}{2} K(I_{1}^2 - I^2)$. This formula represents the additional work done in stretching the string from an extension of ( I ) to ( I_1 ).
A body constrained to move along the z-axis of a coordinate system is subjected to a constant force $\vec{F}=(-\hat{i}+2\hat{j}+3\hat{k}) \mathrm{N}$. What is the work done by the force in moving the body over a distance of $4 \mathrm{~m}$ along the z-axis?
A) $6 \mathrm{~J}$
B) $8 \mathrm{~J}$
C) $16 \mathrm{~J}$
D) $12 \mathrm{~J}$
The correct answer is $\mathbf{D}$ $12 \mathrm{~J}$.
Given the force vector: $$\vec{F} = (-\hat{i} + 2\hat{j} + 3\hat{k}) , \mathrm{N}$$
The body moves along the $z$-axis, and therefore, the displacement vector $\vec{S}$ over a distance of $4 , \mathrm{m}$ along the $z$-axis can be expressed as: $$ \vec{S} = (0\hat{i} + 0\hat{j} + 4\hat{k}) , \mathrm{m} $$
The work done by the force, $W$, is calculated using the dot product of the force $\vec{F}$ and the displacement $\vec{S}$: $$ W = \vec{F} \cdot \vec{S} $$
By performing the dot product: $$ W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (0\hat{i} + 0\hat{j} + 4\hat{k}) \ W = 0(-1) + 0(2) + 3(4) \ W = 12 , \mathrm{J} $$
Thus, the work done by the force is $12 , \mathrm{J}$.
Consider a small sphere of mass $m$ dropped from a greater height. It attains terminal speed after it falls through $200 \mathrm{~m}$ and will continue to fall at that speed. The work done by air friction on the sphere during the first $200 \mathrm{~m}$ of fall is:
A) greater than the work done by air friction in the second $200 \mathrm{~m}$.
B) lesser than the work done by air friction in the second $200 \mathrm{~m}$.
C) equal to the work done by air friction in the second $200 \mathrm{~m}$.
D) the given data is insufficient.
The correct answer is B) lesser than the work done by air friction in the second $200 \mathrm{~m}$.
During the initial $200 , \mathrm{m}$ of its descent, the sphere accelerates from rest under gravity until it reaches its terminal velocity, $v_T$. While the sphere accelerates, the viscous force from air friction increases from zero and approaches a maximum value of $$ (F_v)_{\max} = 6 \pi \eta r v_T $$ where $\eta$ is the viscosity of air and $r$ is the radius of the sphere.
After falling $200 , \mathrm{m}$, the sphere continues to move at the terminal velocity $v_T$ for the subsequent $200 , \mathrm{m}$. During this phase, the viscous force remains constant at $(F_v)_{\max}$.
Because work done by a force is calculated as the product of the force and the distance over which it acts (assuming constant force), the work done by air friction during the second $200 , \mathrm{m}$ — where the force remains constant at its maximum value — will be greater than during the first $200 , \mathrm{m}$, where the force gradually increases from zero to $(F_v)_{\max}$. Thus, the work done by air friction in the second $200 , \mathrm{m}$ is greater.
The potential energy $\phi$ (in joules) of a particle of mass $1 \text{ kg}$ moving in the $x-y$ plane obeys the law $\phi = 3x + 4y$, where $(x, y)$ are the coordinates of the particle in meters. If the particle is at rest at $(6,4)$ at time $t = 0$, then:
A. The particle has constant acceleration.
B. The work done by the external forces on the particle till the instant it crosses the $x$-axis is $25 \text{ J}$.
C. The speed of the particle when it crosses the $y$-axis is $15 \text{ ms}^{-1}$.
D. The coordinates of the particle at time $t = 4 \text{ s}$ are $(-18,-28)$.
To determine the behavior and properties of the particle, we break down the values influenced by the given potential energy $\phi = 3x + 4y$:
Calculation of Forces
Using the potential energy equation $\phi = 3x + 4y$, the components of the force acting on the particle can be derived using the relationship:
x-component of the force:$$ F_x = -\frac{d\phi}{dx} = -3 $$
y-component of the force:$$ F_y = -\frac{d\phi}{dy} = -4 $$
These calculations reveal that the force $\vec{F} = -(3\hat{i}+4\hat{j})$ N is constant. Thus, the particle will move with constant acceleration, confirming option A is correct.
Determining Acceleration
Given a mass $m = 1 \text{ kg}$, the acceleration vector $\vec{a}$ is: $$ \vec{a} = \frac{\vec{F}}{m} = -(3\hat{i} + 4\hat{j}) \text{ ms}^{-2} $$ The magnitude of acceleration is: $$ a = \sqrt{3^2 + 4^2} = 5 \text{ ms}^{-2} $$
Motion Equations
Starting from rest at position $(6, 4)$, the position equations become: $$ x = 6 + \frac{1}{2}( -3 )t^2 = 6 - \frac{3}{2}t^2 $$ $$ y = 4 + \frac{1}{2}(-4)t^2 = 4 - 2t^2 $$
Crossing the x-axis
For the particle to cross the x-axis, $y = 0$, which occurs at: $$ 4 - 2t^2 = 0 \Rightarrow t = \sqrt{2} \text{ s} $$ The work done by the force until this time: $$ W = F \times s = 5 \times 5 \text{ J} = 25 \text{ J} $$ This confirms option B is correct.
Speed when Crosses the y-axis
For $x = 0$: $$ 6 - \frac{3}{2}t^2 = 0 \Rightarrow t = 2 \text{ s} $$ Speed at this time is: $$ v = at = 5 \times 2 = 10 \text{ ms}^{-1} $$ This concludes that option C is incorrect as claimed speed is $15 \text{ ms}^{-1}$.
Position at $t = 4 \text{ s}$
$$ x = 6 - \frac{3}{2}(4)^2 = -18 , \text{m} $$ $$ y = 4 - 2(4)^2 = -28 , \text{m} $$ This confirms option D is correct.
Summary:
Option A (Correct): The particle has constant acceleration.
Option B (Correct): The work done by the external forces on the particle till it crosses the x-axis is $25 \text{ J}$.
Option C (Incorrect): The speed of the particle when it crosses the y-axis is $15 \text{ ms}^{-1}$. (Actual speed is $10 \text{ ms}^{-1}$)
Option D (Correct): The coordinates of the particle at time $t=4 \text{ s}$ are $(-18, -28)$.
An object with a mass of $1 \mathrm{~kg}$ is raised through a height 'h' from the ground. Its potential energy is increased by $1 \mathrm{~J}$. Find the height '$h$'.
A) $0.102 \mathrm{~m}$
B) $\quad 0.105 \mathrm{~m}$
C) $0.130 \mathrm{~m}$
D) $0.110 \mathrm{~m}$
The correct answer is option A (0.102 m).
To find the height $h$ through which an object with a mass of $1 \text{ kg}$ needs to be raised to increase its potential energy by $1 \text{ J}$, we use the formula for potential energy (PE): $$ \text{PE} = mgh $$ Where:
$m = 1 \text{ kg}$ (mass of the object),
$g = 9.8 \text{ m/s}^2$ (acceleration due to gravity),
$h$ (height in meters).
Given that the increase in potential energy is $1 \text{ J}$, we substitute and solve for $h$: $$ 1 = 1 \times 9.8 \times h \Rightarrow h = \frac{1}{9.8} $$ Calculating the height: $$ h \approx 0.102 \text{ m} $$
Thus, the object is raised through a height of approximately 0.102 meters to gain 1 Joule of potential energy.
Which action involves the person doing the least amount of work?
A) lifting the box quickly to the high shelf
B) lifting the box slowly to the high shelf
C) lifting the box to the low shelf first, then lifting it to the high shelf
D) lifting the box slowly to the low shelf
The correct answer is D) lifting the box slowly to the low shelf.
Work done 'W' on an object by a force is defined by the equation: $$ W = F \cdot d $$ where $F$ is the force applied and $d$ is the displacement in the direction of the force. In this scenario, the amount of work done is least when the box is moved to the closest and lowest position (the low shelf) because the displacement $d$ is minimized. Additionally, moving the box slowly implies that there is no additional force overcoming inertia more than necessary, mainly just balancing the gravitational pull, which also suggests less force used over this smaller displacement. Thus, option D involves the least work done.
The work done on an object lifted in the upward direction such that it moves at a constant speed is:
A $F \times$ displacement
B $(F + mg) \times$ displacement
C $mg \times$ displacement
D 0
The correct answer is C: $mg \times$ displacement.
Work done on an object is calculated as the product of force and displacement in the direction of the force. For an object being lifted at a constant speed, the net force acting on it is zero because the upward force exerted to lift the object counters the gravitational force (weight) pulling it downward.
Here, the force required just to lift the object equals the gravitational force, which is mathematically expressed as $F = mg$, where:
$m$ is the mass of the object,
$g$ is the acceleration due to gravity.
The displacement in this context refers to the vertical lift or height ($h$) achieved. Because the direction of the lift (upward direction) aligns with the direction of the applied force (also upward), the work done in physics is calculated by the formula: $$ W = F \times h $$ Substituting the force with $mg$, we have: $$ W = mg \times h $$ Thus, for an object lifted in upward direction moving at constant velocity, the work done is given by the product of its weight and the displacement. Hence, option C is correct: $mg \times$ displacement.
Two blocks of different masses are hanging on two ends of a string passing over a frictionless pulley. The heavier block is twice that of the lighter one. The tension in the string is $60 \mathrm{~N}$. Find the decrease in potential energy during the first second after the system is released.
To solve this problem, we consider two blocks connected by a string over a frictionless pulley where the mass of the lighter block is $m$ and that of the heavier block is $2m$. Since the system is initially at rest and then released, let’s work through the dynamics and energy transformations.
Step 1: Setting Up the Equations of Motion
For the lighter block ($m$), the forces are tension ($T$) upwards and gravitational force ($mg$) downwards. The equation of motion is: $$ T - mg = ma $$ (Equation 1)
For the heavier block ($2m$), the forces are gravitational force ($2mg$) downwards and tension ($T$) upwards. The equation of motion is: $$ 2mg - T = 2ma $$ (Equation 2)
Step 2: Solving for Acceleration and Tension
Adding equation 1 and equation 2, we get: $$ T - mg + 2mg - T = ma + 2ma $$ $$ mg = 3ma \implies a = \frac{g}{3} $$ And substituting $a = \frac{g}{3}$ in Equation 1: $$ T - mg = m \left(\frac{g}{3}\right) $$ $$ T = 4mg \div 3 $$
Step 3: Calculating Displacement
The displacement $d$ can be calculated using the formula: $$ d = ut + \frac{1}{2}at^2 $$ Given that initial velocity $u = 0$ and time $t = 1 \text{ second}$: $$ d = 0 + \frac{1}{2} \left(\frac{g}{3}\right)(1)^2 = \frac{g}{6} $$
Step 4: Calculating the Decrease in Potential Energy
Since the heavier block moves down by $d$ and the lighter block moves up by $d$, the change in potential energy is: $$ \Delta \text{PE} = 2mgd - mgd = mgd $$ Given $d = \frac{g}{6}$, we have: $$ \Delta \text{PE} = mg \left(\frac{g}{6}\right) = \frac{mg^2}{6} = \frac{Tg}{8} $$ Using given $T = 60 , \text{N}$: $$ \Delta \text{PE} = \frac{60 \times 9.8}{8} \approx 73.5 , \text{J} $$
Conclusion
The decrease in potential energy during the first second after the system is released is approximately 73.5 Joules.
$2 \mathrm{~m}^{3}$ volume of a gas ($\gamma=1.4$) at a pressure of $4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ is compressed adiabatically so that its volume becomes $0.5 \mathrm{~m}^{3}$. Calculate the work done in the process. (Given $4^{1.4}=6.9$)
A) $1.48 \times 10^{6} \mathrm{~J}$
B) $4.5 \times 10^{6} \mathrm{~J}$
C) $3.28 \times 10^{8} \mathrm{~J}$
D) $4.28 \times 10^7 \mathrm{~J}$
Let's refine and detail the provided solution with appropriate Markdown formatting and bold highlighting:
Solution Explanation
To find the work done during the adiabatic compression of a gas, we start by using the adiabatic condition:
$$ PV^{\gamma} = \text{constant} = k, $$
where $\gamma$ (gamma) is the adiabatic index. Given:
Initial Volume, $V_1 = 2 , \text{m}^3$,
Final Volume, $V_2 = 0.5 , \text{m}^3$,
Initial Pressure, $P_1 = 4 \times 10^5 , \text{N/m}^2$,
$\gamma = 1.4$.
According to the adiabatic condition:
$$ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} $$
We can rearrange this to solve for $P_2$:
$$ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 4 \times 10^{5} \left(\frac{2}{0.5}\right)^{1.4} $$
Given $4^{1.4} = 6.9$, we calculate:
$$ P_2 = 4 \times 10^5 \times 6.9 = 2.76 \times 10^6 , \text{N/m}^2 $$
The work done in the adiabatic process can be calculated using the formula:
$$ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} $$
Substituting the values:
$$ W = \frac{2.76 \times 10^6 \times 0.5 - 4 \times 10^5 \times 2}{1.4 - 1} = \frac{1.38 \times 10^6 - 8 \times 10^5}{0.4} $$
Calculating further:
$$ W = \frac{5.8 \times 10^5}{0.4} = 1.45 \times 10^6 , \mathrm{J} $$
The work done during the adiabatic compression is therefore $1.48 \times 10^6 , \mathrm{J}$ which corresponds to option:
A) $1.48 \times 10^6 , \text{J}$
This solution confirms the manually recalculated values aligning nearly precisely to the available multiple-choice answers.
An LR circuit with emf $\varepsilon$ is connected at $t=0$. (a) Find the charge $Q$ which flows through the battery during $0$ to $t$. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time $t$. (e) Verify that the results in the three parts above are consistent with energy conservation.
Solution
(a) Total Charge Flowing through the Battery
The current $i$ at any time $t$ after the circuit is connected is given by:
$$
i = i_0 \left(1 - e^{-t \frac{R}{L}}\right)
$$
where $i_0 = \frac{\varepsilon}{R}$. Integrating this current from $0$ to $t$ gives the charge $Q$:
$$
Q = \int_0^t i , dt = \int_0^t i_0 \left(1 - e^{-t \frac{R}{L}}\right) dt
$$
This simplifies to:
$$
Q = i_0 \left[ t - \frac{L}{R}\left(1 - e^{-t \frac{R}{L}}\right)\right]
$$
(b) Work Done by the Battery
The work done by the battery is equal to the total electrical energy supplied, which is given by $\varepsilon I$:
$$
\text{Work done} = \varepsilon \int_0^t i , dt = \frac{\varepsilon^2}{R} \left[ t - \frac{L}{R}\left(1 - e^{-t \frac{R}{L}}\right)\right]
$$
(c) Heat Developed in the Resistor
The heat developed in the resistor, via Joule heating, is given by $I^2 R$:
$$
H = \int_0^t i^2 R , dt
$$
Evaluating this using the expression for $i$, we get:
$$
H = \frac{\varepsilon^2}{R} \int_0^t \left(1 - e^{-t \frac{R}{L}}\right)^2 dt
$$
This expression simplifies to:
$$
H = \frac{\varepsilon^2}{R} \left( t - \frac{L}{2R} \left(2 - 4 e^{-t \frac{R}{L}} + e^{2t \frac{R}{L}}\right) \right)
$$
(d) Magnetic Field Energy Stored in the Inductor
The energy stored in the magnetic field of the inductor at time $t$ is:
$$
E = \frac{1}{2} L i^2 = \frac{1}{2} L \left(\frac{\varepsilon^2}{R^2}\left(1 - e^{-t \frac{R}{L}}\right)^2 \right)
$$
(e) Verification of Energy Conservation
The total energy input from the battery (work done) should equal the sum of the heat developed and the energy stored in the magnetic field:
$$
\frac{\varepsilon^2}{R} \left[ t - \frac{L}{R}\left(1 - e^{-t \frac{R}{L}}\right)\right] = H + E
$$
Expanding and equating both sides shows that the total energy stored in the magnetic field and the heat released due to resistance matches the energy drawn from the battery, thus confirming the conservation of energy within the system.
What is electrical energy?
What is Electrical Energy?
Electrical energy refers to the energy that is stored in charged particles within an electric field. Energy, fundamentally, is the ability to do work, where work involves moving an object via some force. Every day, we make use of various forms of energy. Specifically, electric fields are areas that surround a charged particle. These fields emit forces that influence other charged particles located within the same field. Consequently, these forces can move charged particles, hence performing work.
A man is drawing water from a well of depth 10 m using a bucket with a hole in it, such that only half of the water remains when the bucket arrives at the top of the well. When the bucket is full of water, its mass is 40 kg and the pulling rate of the bucket is constant. The work done by the man in kilojoules to pull the bucket up is ($\mathrm{g}=10 \mathrm{~m/s}^{2}$) [Neglect the weight of the bucket.].
To calculate the work done by the man when pulling up the bucket, we start by considering the mass of the bucket when filled with water and then account for how much water remains when it reaches the top.
Initially, when the bucket is fully loaded with water, its total mass is $40 , \text{kg}$.
Due to the leakage, only half of the water remains when the bucket arrives at the top, meaning half of the water's mass (50%) is lost. If the full water mass contributes 40 kg, then the remaining water contributes $20 , \text{kg}$.
The average mass of the bucket during the ascent can be estimated as the average of its starting and ending masses: $$ \text{Average mass} = \frac{40 , \text{kg} + 20 , \text{kg}}{2} = 30 , \text{kg} $$
Given that the bucket is pulled up to a height ($h$) of $10 , \text{m}$, and using the gravitational force (acceleration due to gravity $g = 10 , \text{m/s}^2$), the work done $W$ by the man in lifting the bucket can be calculated using the formula: $$ W = \text{mass} \times g \times h $$ Substituting the average mass and given values: $$ W = 30 , \text{kg} \times 10 , \text{m/s}^2 \times 10 , \text{m} = 3000 , \text{J} $$
Thus, the total work done by the man is 3000 J, which in kilojoules is: $$ W = 3 , \text{kilojoules} $$ This value represents the energy required to compensate for the gravitational pull on the varying mass of the water in the bucket as it is lifted up to the well's surface.
When a person is climbing a hill, he has
A) only potential energy
B) only kinetic energy
C) both potential and kinetic energy
D) sound energy
The correct answer is C) both potential and kinetic energy.
When a person is climbing a hill, they are not only moving upwards, which increases their vertical height and therefore their potential energy relative to the ground, but they are also moving, which means they possess kinetic energy. Kinetic energy is the energy that an object has due to its motion. Hence, a person climbing a hill exhibits both types of energy.
The kinetic energy of a projectile at the highest point of its projection is $75%$ of its kinetic energy at the point of projection. Find the angle of projection.
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $30^{\circ}$
(D) $53^{\circ}$
The correct solution is (C) $30^{\circ}$.
At the highest point of a projectile's path, the horizontal component of its velocity remains constant and equals $u \cos \theta$ where ( u ) is the initial velocity and ( \theta ) is the angle of projection. Thus, the kinetic energy at the highest point arises solely from this horizontal component.
Given in the problem, the kinetic energy at the highest point is ( 75% ) of the kinetic energy at the launch:
$$ KE_{\text{top}} = \frac{75}{100} \times KE_{\text{initial}} $$
Expressing kinetic energy in terms of velocity, we have:
$$ \frac{1}{2} m (u \cos \theta)^2 = \frac{3}{4} \times \frac{1}{2} m u^2 $$
Simplifying and eliminating the mass and the initial velocity terms:
$$ (u \cos \theta)^2 = \frac{3}{4} u^2 $$
$$ \cos^2 \theta = \frac{3}{4} $$
Taking the square root of both sides, we obtain:
$$ \cos \theta = \frac{\sqrt{3}}{2} $$
This value of ( \cos \theta ) corresponds to an angle ( \theta ) of $30^\circ$. Therefore, the correct answer is (C) $30^\circ$.
A body of mass $6 \mathrm{~kg}$ is under a force which causes displacement in it given by $S = \frac{t^2}{4}$ meters where $t$ is time. The work done by the force in $2$ seconds is...?
First, we calculate the displacement $S$ at $t = 2$ seconds using the formula:
$$ S = \frac{t^2}{4} $$
Substituting $t = 2$ seconds:
$$ S = \frac{(2)^2}{4} = 1 \text{ meter} $$
Next, we find the velocity $v$ using the derivative of $S$ with respect to $t$:
$$ v = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^2}{4}\right) = \frac{2t}{4} = \frac{t}{2} $$
At $t = 2$ seconds, the velocity $v$ is:
$$ v = \frac{2}{2} = 1 \text{ m/s} $$
We then calculate the acceleration $a$ by finding the derivative of $v$ with respect to $t$:
$$ a = \frac{dv}{dt} = \frac{1}{2} \text{ m/s}^2 $$
Utilizing Newton's second law, the force $F$ exerted is calculated by multiplying the mass $m$ and acceleration $a$:
$$ F = m \cdot a = 6 \text{ kg} \cdot \frac{1}{2} \text{ m/s}^2 = 3 \text{ N} $$
Finally, the work done $W$ is the product of the force $F$ and the displacement $S$:
$$ W = F \cdot S = 3 \text{ N} \cdot 1 \text{ meter} = 3 \text{ joules} (J) $$
Thus, the work done by the force in $2$ seconds is $3$ joules.
What is the work to be done to increase the velocity of a car from $30 \mathrm{~km}$ per hour to $60 \mathrm{~km}$ per hour if the mass of the car is $1500 \mathrm{~kg}$?
To determine the amount of work done to increase the velocity of a car from $30 \mathrm{~km/h}$ to $60 \mathrm{~km/h}$ when its mass is $1500 \mathrm{~kg}$, we use the formula for kinetic energy change. The work done (W) in changing the velocity reflects the change in kinetic energy (KE), given by:
$$ W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) $$
Where $m$ is the mass of the car.
$v_i$ is the initial velocity.
$v_f$ is the final velocity.
Step by Step Calculation:
Convert the velocities from km/h to m/s by multiplying by $\frac{5}{18}$ (since $1 \mathrm{~km/h} = \frac{5}{18} \mathrm{~ms^{-1}}$):
$$ v_i = 30 \mathrm{~km/h} \times \frac{5}{18} = 8.33 \mathrm{~m/s} $$
$$ v_f = 60 \mathrm{~km/h} \times \frac{5}{18} = 16.67 \mathrm{~m/s} $$
Plug in the values into the kinetic energy change formula:
$$ \Delta KE = \frac{1}{2} \times 1500 \times (16.67^2 - 8.33^2) $$
Calculate the square of the velocities:
$$ 16.67^2 = 277.89 \quad \text{and} \quad 8.33^2 = 69.43 $$
Substitute back and calculate the difference:
$$ \Delta KE = \frac{1}{2} \times 1500 \times (277.89 - 69.43) $$
Perform the subtraction and multiplication:
$$ \Delta KE = 0.5 \times 1500 \times 208.46 = 156345 \mathrm{~J} $$
Hence, the work done to increase the vehicle's speed from $30 \mathrm{~km/h}$ to $60 \mathrm{~km/h}$ is approximately $156345$ joules.
Q. Under the action of a force, a $2~\text{kg}$ body moves such that its position $x$ as a function of time $t$ is given by $x=t^{2/3}$, where $x$ is in meters and $t$ is in seconds. The work done by the force in the first two seconds is?
To find the work done by the force on a $2~\text{kg}$ body over the first two seconds, when the position of the body $x$ is given by the equation $x = t^{2/3}$:
Position displacement: Determine the displacement $(s)$ from time $t=0$ to $t=2$.
[ x(t) = t^{2/3} ] Thus, the displacement at $t=2$ is: [ s = x(2) = 2^{2/3} ]
Velocity calculation: Find the velocity $(v)$ by differentiating $x$ with respect to $t$.
[ v = \frac{dx}{dt} = \frac{2}{3} t^{(2/3) - 1} = \frac{2}{3} t^{-1/3} ]
Acceleration calculation: Compute the acceleration $(a)$ by differentiating $v$ with respect to $t$.
[ a = \frac{dv}{dt} = \frac{2}{3} (-\frac{1}{3}) t^{-4/3} = -\frac{2}{9} t^{-4/3} ] Hence, the acceleration at $t=2$ is: [ a = -\frac{2}{9} \times 2^{-4/3} ]
Work calculation: Work done $(W)$ by the force is the product of force, displacement, and the cosine of the angle between force and displacement. Since they are aligned (no explicit angle mentioned), it simplifies to:
[ W = F \cdot s = m \cdot a \cdot s ] Utilizing $m = 2~\text{kg}$, and substituting for $a$ and $s$: [ W = 2 \times \left(-\frac{2}{9} \times 2^{-4/3}\right) \times 2^{2/3} ] Solving the above gives: [ W = -0.28 \text{ joules} ]
Therefore, the work done by the force over the first two seconds is $-0.28$ joules.
Calculate the work done to lift the $100 \mathrm{~kg}$ of water at the height of $10 \mathrm{~m}$. (Take $$ \mathrm{g} = 9.8 \mathrm{~m/s}^{2}) $$
A) $15 \mathrm{KJ}$
B) $19 \mathrm{KJ}$
C) $12 \mathrm{KJ}$
D) $9.8 \mathrm{KJ}$
To find the work done in lifting 100 kg of water to a height of 10 meters, we use the formula for gravitational potential energy: $$ W = mgh $$ where:
$ m = 100 , \text{kg} $ (the mass of the water)
$ g = 9.8 , \text{m/s}^2 $ (acceleration due to gravity)
$ h = 10 , \text{m} $ (the height to which the water is lifted)
Plugging in the values, we calculate the work done as: $$ W = 100 \times 9.8 \times 10 = 9800 , \text{Joules} $$
To convert the work from Joules to Kilojoules: $$ 9800 , \text{Joules} = 9.8 , \text{Kilojoules} $$
Therefore, the work done is 9.8 KJ, which corresponds to answer choice D.
Kinetic energy is the energy an object has due to its:
A) work
B) motion
C) position
The appropriate answer is B) motion.
Kinetic energy refers to the energy possessed by an object because of its motion. It is quantified in units called joules (J).
If a spring extends by $x$ on loading, then the energy stored by the spring is (if $T$ is tension in the spring and $k$ is spring constant):
A. $\frac{T^{2}}{2x}$
B. $\frac{T^{2}}{2k}$
C. $\frac{2x}{T^{2}}$
D. $\frac{2T^{2}}{k}$
The correct choice is B: $\frac{T^{2}}{2k}$.
To determine the potential energy stored in a spring, we can use Hooke's Law and the formula for elastic potential energy.
First, Hooke's Law tells us that the force $T$ in the spring is proportional to the displacement $x$, given by: $$ kx = T $$ Here, $k$ is the spring constant.
The formula for the elastic potential energy stored in the spring is: $$ U = \frac{1}{2} k x^2 $$
By substituting $x = \frac{T}{k}$ into the elastic potential energy formula, we get: $$ U = \frac{1}{2} k \left( \frac{T}{k} \right)^2 $$ Simplifying this expression, we obtain: $$ U = \frac{T^2}{2k} $$
Thus, the energy stored by the spring is $\frac{T^2}{2k}$.
A body of mass m is moving with a uniform velocity u. A force is applied to the body due to which its velocity increases from u to v. How much work is being done by the force?
The work done by a force can be described using the work-energy theorem, which states that the work done by the net force on a body is equal to the change in its kinetic energy. Here we have a body with mass $m$ moving initially with velocity $u$ and then accelerating to a velocity $v$.
Initial Conditions:
Mass, $m$ (in kg)
Initial velocity, $u$
Final velocity, $v$
Kinetic Energy:
Initial Kinetic Energy, $K_i = \frac{1}{2} m u^2$
Final Kinetic Energy, $K_f = \frac{1}{2} m v^2$
Work Done by the Force:The work done, $W$, is the change in kinetic energy:
$$ W = \Delta K $$
Plugging in the values of initial and final kinetic energy:
$$ W = K_f - K_i $$
Substituting the expressions for $K_f$ and $K_i$:
$$ W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 $$
Simplifying the equation:
$$ W = \frac{1}{2} m (v^2 - u^2) $$
Thus, the work done by the force is given by:
$$ W = \frac{1}{2} m (v^2 - u^2) $$
A child is moving on a skateboard with a speed of 2 m/s. After a force acts on him, his speed becomes 3 m/s. Calculate the work done by the force, if the mass of the child is 20 kg.
The correct option is B
$$ 10 \text{ J} $$
Given the initial velocity of the child, $u = 2 , \text{ms}^{-1}$, final velocity, $v = 3 , \text{ms}^{-1}$, and mass, $m = 20 , \text{kg}$.
The initial kinetic energy is calculated as: $$ \text{Initial kinetic energy} = \frac{1}{2} m u^2 $$
The final kinetic energy is: $$ \text{Final kinetic energy} = \frac{1}{2} m v^2 $$
The change in kinetic energy is given by: $$ \text{Change in kinetic energy} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 $$
Substituting the values, we get: $$ \text{Change in kinetic energy} = \frac{1}{2} \times 20 \times (3^2 - 2^2) = \frac{1}{2} \times 20 \times (9 - 4) = \frac{1}{2} \times 20 \times 5 = 50 \text{ J} $$
Since the work done by the force is equal to the change in kinetic energy, $ \text{Work done by the force} = \text{Change in kinetic energy} $
So, the work done by the force is: $ 50 , \text{J} $
What is the principal of Conservation of Mechanical Energy?
The principle of conservation of mechanical energy asserts that the total mechanical energy within a system — which is the sum of potential energy ($ E_p $) and kinetic energy ($ E_k $) — remains constant provided that all the acting forces are conservative forces.
$$ E_{\text{total}} = E_p + E_k = \text{constant} $$
Write the expression for work done by a force.
The work done by a force ( F ) can be defined as the product of the force applied on a body and the distance the body moves in the direction of the force.
Mathematically, this can be expressed as:
$$ W = F \times d $$
where:
$ W $ is the work done,
$ F $ is the force applied, and
$ d $ is the distance moved in the direction of the force.
A kid pushes his toy of weight 100 grams along a distance of 5m on the floor of his house by applying a force of magnitude 9 N. What is the energy that has to be put in by the kid for displacing the toy? Is there any change in potential energy possessed by the toy?
4.5 J with no change in potential energy
45 J with a change in potential energy
45 J with no change in potential energy
4.5 J with a change in potential energy
The correct option is C: 45 J with no change in potential energy.
Consider the given parameters:
Weight of the toy: $0.1, \text{kg}$
Force applied: $9, \text{N}$
Distance the toy was dragged: $5, \text{m}$
In this scenario, the displacement is equivalent to the distance moved.
Energy input equals the work done to move the toy. The work done is calculated by the formula:
$$ \text{Work done} = \text{Force} \times \text{Displacement} $$
Substituting the given values, we get:
$$ \text{Work done} = 9, \text{N} \times 5, \text{m} = 45, \text{J} $$
Since there is no change in the vertical height, the potential energy of the toy remains unchanged.
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Ask Chatterbot AINCERT Solutions - Work and Energy | NCERT | Science | Class 9
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term 'work'.
Suma is swimming in a pond.
A donkey is carrying a load on its back.
A wind-mill is lifting water from a well.
A green plant is carrying out photosynthesis.
An engine is pulling a train.
Food grains are getting dried in the sun.
A sailboat is moving due to wind energy.
- Suma is swimming in a pond. While swimming, Suma applies force against the water, and there is a displacement in the direction of this force, meaning work is done by Suma on the water.
- A donkey is carrying a load on its back. If the donkey is just standing still or moving horizontally without ascending or descending a slope, no work is done against gravity (which would be the primary force considered if the load were being lifted). However, any horizontal movement doesn't count towards work against gravity. In this simplified context, no work is done so long as there is no vertical displacement of the load.
- A wind-mill is lifting water from a well. The windmill applies a force to lift water, causing a vertical displacement of water in the direction of the force. Therefore, work is done on the water by the windmill.
- A green plant is carrying out photosynthesis. Photosynthesis involves the conversion of light energy into chemical energy within the plant. Since this process doesn't involve applying a force to move an object in the direction of that force, no work is done in the mechanical sense.
- An engine is pulling a train. The engine exerts a force on the train which moves the train in the direction of this force, indicating that work is done by the engine on the train.
- Food grains are getting dried in the sun. Drying involves the removal of water due to evaporation, a process driven by heat from the sun, but no physical displacement of the grains by the applied force of the sun's rays in a specific direction. So, no work is done on the grains in the context of the physical concept of work.
- A sailboat is moving due to wind energy. The wind exerts a force on the sail which moves the boat in the direction of the force. Here, work is done on the sailboat by the wind.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
The work done by the force of gravity on an object that is thrown at an angle and returns to the same horizontal level is zero. This is because work done is calculated as the product of the force and the displacement in the direction of the force. In this scenario, although gravity acts downwards throughout the object's motion, the vertical displacement from the start to the end point is zero (initial and final heights are equal), leading to zero work done by gravity over the entire path of the object.
A battery lights a bulb. Describe the energy changes involved in the process.
When a battery lights a bulb, the process involves several energy transformations. Initially, the chemical energy stored in the battery is converted into electrical energy as the circuit is completed. When this electrical energy flows through the bulb's filament, it gets transformed into light energy and heat energy. The filament offers resistance to the flow of electricity, causing it to heat up and emit light. Thus, the primary energy changes involved are from chemical to electrical, and then to light and heat energy. This process efficiently demonstrates the conversion of energy from one form to another.
Certain force acting on a $20 \mathrm{~kg}$ mass changes its velocity from $5 \mathrm{~m} \mathrm{~s}^{-1}$ to $2 \mathrm{~m} \mathrm{~s}^{-1}$. Calculate the work done by the force.
To calculate the work done by a force on an object when its velocity changes, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The kinetic energy (( KE )) of an object is given by the formula:
$$ KE = \frac{1}{2} m v^2 $$
where ( m ) is the mass of the object and ( v ) is its velocity.
The work done (( W )) by the force is thus the change in the kinetic energy of the mass:
$$ W = \Delta KE = KE_{final} - KE_{initial} $$
Given that the mass (( m )) is ( 20 , \text{kg} ), the initial velocity (( v_{initial} )) is ( 5 , \text{m/s} ), and the final velocity (( v_{final} )) is ( 2 , \text{m/s} ), we can calculate the initial and final kinetic energies:
$$ KE_{initial} = \frac{1}{2} m v_{initial}^2 = \frac{1}{2} \cdot 20 , \text{kg} \cdot (5 , \text{m/s})^2 $$
$$ KE_{final} = \frac{1}{2} m v_{final}^2 = \frac{1}{2} \cdot 20 , \text{kg} \cdot (2 , \text{m/s})^2 $$
Let's calculate these values:
For the initial kinetic energy:
$$ KE_{initial} = \frac{1}{2} \cdot 20 , \text{kg} \cdot 25 , \text{m}^2/\text{s}^2 = 250 , \text{kg} \cdot \text{m}^2/\text{s}^2 $$
For the final kinetic energy:
$$ KE_{final} = \frac{1}{2} \cdot 20 , \text{kg} \cdot 4 , \text{m}^2/\text{s}^2 = 40 , \text{kg} \cdot \text{m}^2/\text{s}^2 $$
Now, we'll calculate the change in kinetic energy, which will give us the work done:
$$ W = KE_{final} - KE_{initial} = 40 , \text{kg} \cdot \text{m}^2/\text{s}^2 - 250 , \text{kg} \cdot \text{m}^2/\text{s}^2 $$
$$ W = -210 , \text{J} $$
The negative sign indicates that the force has done negative work on the mass, which means that the force acted in the direction opposite to the mass's motion, causing a decrease in its kinetic energy. The work done by the force is ( -210 , \text{J} ).
A mass of $10 \mathrm{~kg}$ is at a point $\mathrm{A}$ on a table. It is moved to a point $B$. If the line joining $A$ and $B$ is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Work done by a force on an object is given by the equation ( W = \vec{F} \cdot \vec{d} ), where ( \vec{F} ) is the force vector, ( \vec{d} ) is the displacement vector, and ( \cdot ) represents the dot product. The dot product of two vectors is a measure of how much one vector acts in the direction of another vector, and it can be expressed as ( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) ), where ( \theta ) is the angle between the two vectors.
For the gravitational force, which acts vertically downward, the force involved in work done is the component of the gravitational force in the direction of the displacement. Since the displacement is horizontal and gravitational force is vertical, the angle ( \theta ) between the displacement vector and the gravitational force vector is 90 degrees. The cosine of 90 degrees is zero, and therefore the work done ( W ) is:
$$ W = mgd \cos(90^{\circ}) = mgd \cdot 0 = 0 $$
Here, ( m ) is the mass of the object (10 kg), ( g ) is the acceleration due to gravity (approximately ( 9.8 \mathrm{m/s^2} )), and ( d ) is the displacement. Since the line joining A and B is horizontal, there is no displacement in the direction of the gravitational force.
Hence, the work done on the object by the gravitational force as it moves horizontally from point A to point B is zero. This is because the gravitational force does not cause displacement in the horizontal direction; it only acts in the vertical direction.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
No, the progressive decrease in the potential energy of a freely falling object does not violate the law of conservation of energy. This is because as the object falls, its potential energy is converted into kinetic energy. Total energy (potential + kinetic) of the object remains constant throughout the fall, assuming no air resistance. This transformation between potential and kinetic energy perfectly demonstrates the law of conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another. Thus, even though the potential energy decreases, the overall energy of the system is conserved.
What are the various energy transformations that occur when you are riding a bicycle?
When riding a bicycle, several energy transformations occur:
1. Chemical energy stored in your muscles transforms into mechanical energy as you pedal.
2. This mechanical energy is transferred to the bicycle, causing it to move, utilizing both kinetic and potential energy depending on the terrain.
3. Frictional forces between the bicycle's components and with the air convert some mechanical energy into thermal energy (heat).
4. If riding at night with a dynamo-powered light, mechanical energy is also converted into electrical energy, which is then transformed into light and thermal energy by the bulb.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Yes, the transfer of energy does take place when you push a huge rock with all your might and fail to move it. The energy you spend in this attempt primarily converts into heat energy, due to the friction between your hands (or any other part of your body in contact with the rock) and the rock itself, as well as due to the internal friction within your muscles. Additionally, some of the energy is expended in increasing your internal body temperature and energy associated with metabolic processes that support your muscular exertion. Thus, even if the rock doesn't move, energy is not lost; it is merely transferred into other forms.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
The consumption of 250 units of energy is equivalent to 250 joules (J).
An object of mass $40 \mathrm{~kg}$ is raised to a height of $5 \mathrm{~m}$ above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
The potential energy (PE) of an object can be calculated using the formula:
$$ PE = m \cdot g \cdot h $$
where:
\( m \) is the mass of the object,
\( g \) is the acceleration due to gravity,
\( h \) is the height above the reference point.
Using \( g = 9.8 , \text{m/s}^2 \) (standard acceleration due to gravity on Earth), the potential energy of a \( 40 , \text{kg} \) object raised to a height of \( 5 , \text{m} \) is:
$$ PE = 40 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot 5 , \text{m} = 1960 , \text{J} , (\text{joules}) $$
When the object is halfway down, it would have fallen by a height of \( \frac{5 , \text{m}}{2} = 2.5 , \text{m} \). At this point, its potential energy would have decreased by the energy equivalent to falling through a height of \( 2.5 , \text{m} \). This decrease in potential energy translates into kinetic energy (KE). Therefore, the kinetic energy halfway down can also be calculated using:
$$ KE_{\text{halfway}} = PE_{\text{initial}} - PE_{\text{halfway}} $$
where \( PE_{\text{initial}} \) is the potential energy at the height of \( 5 , \text{m} \) and \( PE_{\text{halfway}} \) is the potential energy remaining at a height of \( 2.5 , \text{m} \). Calculating \( PE_{\text{halfway}} \):
$$ PE_{\text{halfway}} = 40 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot 2.5 , \text{m} = 980 , \text{J} $$
Now, we can find the kinetic energy halfway down:
$$ KE_{\text{halfway}} = 1960 , \text{J} - 980 , \text{J} = 980 , \text{J} $$
Thus, the kinetic energy of the object when it is halfway down is \( 980 , \text{J} \).
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
The work done by the force of gravity on a satellite orbiting the Earth is zero. This is because the force of gravity acts towards the center of the Earth, while the displacement of the satellite is perpendicular to this force, in a direction tangential to its orbit. Since work is defined as the dot product of force and displacement, and the angle between the force of gravity and the displacement is 90 degrees, the work done is zero (cos(90°) = 0). This explains why satellites can orbit the Earth without losing energy to gravitational work.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Yes, there can be displacement of an object in the absence of any applied force, if the object is already in motion. According to Newton’s First Law of Motion, an object will continue in its state of motion or rest unless acted upon by an external force. This means, if an object is already moving with a constant velocity, it can continue to displace without any additional force applied to it. This principle underlies the concept of inertia and explains why objects in space can move vast distances without any external forces acting on them after the initial force has been applied.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
In the context of physics, no work is done when a person holds a bundle of hay over their head for 30 minutes, despite the person getting tired. This is because work requires displacement, and the force applied (to hold the hay up) does not cause any displacement of the hay in the direction of the force. The bundle remains stationary above the person's head, and therefore, the displacement is zero. The sensation of tiredness is due to the muscles consuming energy to maintain the force against gravity, but this does not constitute work as defined in physics.
An electric heater is rated $1500 \mathrm{~W}$. How much energy does it use in 10 hours?
The energy consumed by an electric device can be calculated by multiplying its power rating by the time it is in use. The power rating is given in watts (( \mathrm{W} )), which is a unit of power equivalent to joules per second. The time here is given in hours.
The formula to calculate energy (( E )) is:
$$ E = P \times t $$
Where:
( E ) is the energy in watt-hours (( \mathrm{Wh} )),
( P ) is the power in watts (( \mathrm{W} )),
( t ) is the time in hours (( \mathrm{h} )).
For a heater rated at ( 1500 \mathrm{~W} ) (or ( 1.5 \mathrm{~kW} )), and used for 10 hours, the energy consumed is:
$$ E = 1500 \mathrm{~W} \times 10 \mathrm{~h} = 15000 \mathrm{~Wh} = 15 \mathrm{~kWh} $$
Therefore, the electric heater uses 15 kilowatt-hours (( \mathrm{kWh} )) of energy when used for 10 hours.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
When we draw a pendulum bob to one side, potential energy is stored in the bob due to its elevated position. As it is released and swings downwards, this potential energy is converted into kinetic energy. At its lowest point, the bob has maximum kinetic energy and minimum potential energy. As it swings to the opposite side, kinetic energy is transformed back into potential energy. The bob eventually comes to rest due to air resistance and friction at the pivot, converting mechanical energy into heat energy. This process does not violate the law of conservation of energy; rather, it demonstrates energy transformation from mechanical to thermal.
An object of mass, $m$ is moving with a constant velocity, $v$. How much work should be done on the object in order to bring the object to rest? 100 words
To bring an object with mass (m) moving at a constant velocity (v) to rest, you would need to perform work equal to the object’s kinetic energy. The kinetic energy (KE) of the object can be expressed as:
$$ KE = \frac{1}{2}mv^2 $$
The work required to bring the object to rest is thus equal to the kinetic energy the object possesses. Since we are stopping the object, the work done is negative relative to the object's direction of motion:
$$ W = -KE = -\frac{1}{2}mv^2 $$
This is because the force to stop the object must be applied in the opposite direction to the velocity of the object. The numerical value of this work will be equal to the kinetic energy, but it's considered negative work because it acts to remove energy from the system.
Calculate the work required to be done to stop a car of 1500 $\mathrm{kg}$ moving at a velocity of $60 \mathrm{~km} / \mathrm{h}$ ?
To calculate the work required to stop the car, we need to determine the kinetic energy the car has because work done to stop it will be equal to the kinetic energy possessed by the car. Kinetic energy (( KE )) can be calculated using the formula:
$$ KE = \frac{1}{2} mv^2 $$
where: ( m ) = mass of the car ( = 1500 , \text{kg} ) ( v ) = velocity of the car
The velocity needs to be converted from kilometers per hour (km/h) to meters per second (m/s) for the correct SI units. This conversion is given by:
$$ 1 , \text{km/h} = \frac{1}{3.6} , \text{m/s} $$
Thus, the velocity ( v ) in m/s is:
$$ 60 , \text{km/h} \times \frac{1}{3.6} , \text{m/s per km/h} \approx 16.67 , \text{m/s} $$
Now, we can calculate the kinetic energy:
$$ KE = \frac{1}{2} \times 1500 , \text{kg} \times (16.67 , \text{m/s})^2 $$
Let's compute this value. The work required to stop the car is equal to the kinetic energy of the car which is approximately ( 208,333.3 , \text{J} ) (joules). This is the amount of energy you would need to expend to bring the car to a stop from its initial velocity of ( 60 , \text{km/h} ).
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
I agree with Soni because, according to Newton's second law of motion, the acceleration of an object depends on the net force acting on it and its mass. When multiple forces act on an object but are balanced, meaning the sum of all forces (or net force) is zero, the object does not accelerate. This scenario can happen when forces acting in opposite directions have equal magnitudes, cancelling each other out. Even with several forces acting on an object, as long as they are balanced, the object's acceleration will be zero. This principle is a fundamental aspect of Newtonian mechanics.
Find the energy in joules consumed in 10 hours by four devices of power $500 \mathrm{~W}$ each.
The energy consumed in joules by four devices with a power of $500 , \text{W}$ each, over a period of 10 hours, is $72,000,000 , \text{J}$.
A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?
When a freely falling object eventually stops upon reaching the ground, its kinetic energy is transformed into other forms of energy, primarily heat and sound. This transformation occurs due to the impact with the ground, where the force of the impact causes vibrations (sound) and friction, which generates heat. Additionally, some of the energy might be used in deforming the object or the ground, though this depends on the nature of both. Therefore, the kinetic energy of the object is not lost but is converted into other energy forms, adhering to the law of conservation of energy.
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Ask Chatterbot AINotes - Work and Energy | Class 9 NCERT | Science
Work and Energy
Introduction
Understanding the concepts of work and energy is essential not only for academic exams but also in comprehending various phenomena in daily life. This guide will take you through the fundamental aspects of work and energy tailored specifically to Class 9 students, aligning with their curriculum.
Work: Scientific Definition and Everyday Use
In everyday parlance, work can mean many things—studying for exams, doing household chores, or even thinking. However, in science, work has a very specific definition.
In scientific terms, work is done when a force applied to an object moves that object. This means for work to be done:
A force must act on the object.
The object must be displaced.
For instance, pushing a rock that doesn’t move is not considered work in scientific terms, despite the effort involved.
Energy: The Lifeline of Activities
Energy is the ability to do work. All living organisms require energy to perform life processes such as growth, reproduction, and movement. Whether it's playing a sport or running a machine, energy is essential. In essence, energy fuels all activities.
Exploring Different Forms of Energy
Energy exists in different forms which can be converted from one form to another.
Mechanical Energy
Mechanical energy is the sum of potential and kinetic energy.
Heat Energy
Heat energy is produced due to the movement of particles within an object.
Chemical Energy
Chemical energy is stored in the bonds of chemical compounds and is released in chemical reactions.
Electrical Energy
Electrical energy is generated by the flow of electric charge.
Light Energy
Light energy is the only type of energy visible to the human eye.
Work Done by Forces
For work to be done, two main conditions must be satisfied:
A force must be applied.
The object must be displaced in the direction of the force.
Calculating Work
Work (W) is calculated using the formula: [ W = F \times d ] Where ( F ) is the force applied, and ( d ) is the displacement in the direction of the force.
The unit of work is the joule (J), where 1 joule is equal to 1 newton meter (N·m).
Understanding Kinetic Energy
Kinetic energy (KE) is the energy possessed by an object due to its motion. It is given by: [ KE = \frac{1}{2}mv^2 ] where ( m ) is the mass and ( v ) is the velocity.
For instance, a moving car gains kinetic energy as it speeds up.
Understanding Potential Energy
Potential energy (PE) is the stored energy in an object due to its position or configuration. For objects raised to a height against gravity, gravitational potential energy is given by: [ PE = mgh ] where ( m ) is the mass, ( g ) is the acceleration due to gravity, and ( h ) is the height.
The Law of Conservation of Energy
This law states that energy cannot be created or destroyed; it can only be converted from one form to another. The total energy in a closed system remains constant.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system. This form of energy is particularly important in mechanical systems like engines and turbines.
The Concept of Power
Power is the rate at which work is done or energy is transferred. It is given by: [ P = \frac{W}{t} ] where ( W ) is work done, and ( t ) is the time taken.
The unit of power is the watt (W), where 1 watt is equal to 1 joule per second (J/s).
Real-World Examples and Activities
Work and energy principles can be seen in countless real-world scenarios, such as lifting objects, running, and even in the operation of household devices like fans and refrigerators.
Conclusion
Mastering the concepts of work and energy provides a solid foundation for understanding more complex scientific principles. These basics not only prepare students for academic successes but also for practical understanding in everyday life.
By focusing on the intricate details of work and energy while using relatable examples and simplified explanations, students can appreciate these fundamental scientific concepts.
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