Academia Aeternum
तमसो मा ज्योतिर्गमय
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Q1
The sign of work done by a force on a body is important to understand. State carefully
if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform
velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Work done by a force depends on the angle between the force and displacement vectors. If the force has a component along the direction of displacement, it performs work.
(a) Work done by a man lifting a bucket out of a well
The man applies an upward force through the rope and the bucket also moves upward. Thus force and displacement are in the same direction.
\( \theta = 0^\circ \Rightarrow \cos 0 = 1 \)
Work done by the man is Positive.
(b) Work done by gravitational force
Gravity acts downward while the bucket moves upward. Thus force and displacement are in opposite directions.
\( \theta = 180^\circ \Rightarrow \cos 180^\circ = -1 \)
Work done by gravity is Negative.
(c) Work done by friction on a body sliding down an inclined plane
When a body slides down an incline, friction always acts opposite to motion.
Since friction opposes displacement,
\( \theta = 180^\circ \)
Work done by friction is Negative.
(d) Work done by applied force on a body moving with uniform velocity on a rough horizontal plane
Uniform velocity means the applied force balances friction. However, the applied force acts in the direction of motion.
\( \theta = 0^\circ \)
Work done by the applied force is Positive.
(e) Work done by air resistance on a vibrating pendulum
Air resistance always acts opposite to the direction of motion of the pendulum.
Hence the force is opposite to displacement.
\( \theta = 180^\circ \)
Work done by air resistance is Negative.
Q2
A body of mass 2 kg initially at rest moves under the action of an applied
horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,
and interpret your results.
When a body moves on a rough horizontal surface, the applied force must overcome the kinetic friction. The net force produces acceleration according to Newton’s Second Law. Work done by forces during motion changes the kinetic energy of the body.
Given
Q3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
In classical mechanics, the motion of a particle in a potential field is governed by its total mechanical energy:
Since kinetic energy cannot be negative,
\[ K = E - V(x) \ge 0 \] Therefore, \[ E \ge V(x) \]Hence a particle can exist only in regions where the total energy is greater than or equal to the potential energy.
The potential curve is parabolic and symmetric about the origin. This corresponds to the potential of a simple harmonic oscillator.
\[ V(x) = \frac{1}{2}kx^2 \]The particle oscillates between the two turning points.
This potential has two minima separated by a central barrier.
The potential suddenly increases at a certain point.
The particle moves in a finite potential well with walls.
Q4 The potential energy function for a particle executing linear simple harmonic motion is given by \( V(x) = \frac{kx^2}{2} \), where \(k\) is the force constant of the oscillator. For \(k = 0.5\,\text{N m}^{-1}\), the graph of \(V(x)\) versus \(x\) is shown in Fig. 5.12. Show that a particle of total energy \(1\,\text{J}\) moving under this potential must turn back when it reaches \(x = \pm 2\,\text{m}\).
In Simple Harmonic Motion (SHM), the potential energy of a particle depends on displacement as
At the turning points, the particle momentarily stops and reverses its direction of motion.
Given
At the turning point, kinetic energy becomes zero. Therefore,
\[ E = V(x) \]In SHM, the maximum displacement from equilibrium is called the amplitude.
\[ A = \pm 2\,\text{m} \] Thus the particle oscillates between \[ -2\,\text{m} \quad \text{and} \quad +2\,\text{m} \]
Q5
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy
required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to
the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force
over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to
dissipation against atmospheric resistance, however small. Why then does its speed increase
progressively as it comes closer and closer to the earth ?
(d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks
the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at
its other end. In which case is the work done greater ?
Work done by a force depends on the component of the force along the direction of displacement.
Another important principle used here is that gravitational force is a conservative force, meaning the work done over a complete closed path is zero.
The heat required for burning the casing comes from the kinetic energy of the rocket.
Therefore, the heat energy comes at the expense of the rocket’s kinetic energy, not the atmosphere.
The gravitational force between the Sun and the comet is a conservative force.
For conservative forces,
\[ W = \oint \vec{F}\cdot d\vec{r} = 0 \]because the comet returns to the same position after completing the orbit.
Hence the net work done over a complete orbit is zero.
Atmospheric drag removes mechanical energy from the satellite.
As the satellite loses energy, its orbit gradually shrinks and the satellite moves to a lower orbit.
In orbital motion,
\[ v = \sqrt{\frac{GM}{r}} \]Thus when orbital radius \(r\) decreases, velocity increases.
Therefore the satellite moves faster as it comes closer to Earth even though its total energy decreases.
Case (i): Carrying the mass
The force exerted by the man is vertical (supporting the weight), while the displacement is horizontal.
\[ \theta = 90^\circ \] \[ W = Fs\cos90^\circ = 0 \]Thus no work is done on the mass.
Case (ii): Pulling the rope
The man pulls the rope causing the hanging mass to rise.
Tension in the rope equals the weight of the mass.
\[ T = mg \] \[ T = 15 \times 9.8 = 147\,N \]Displacement of the mass = 2 m.
\[ W = T \times s \] \[ W = 147 \times 2 \] \[ W = 294\,J \]Thus work is done against gravity.
Q6
Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of
the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential
energy.
(c) The rate of change of total momentum of a many-particle system is proportional
to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after
the collision are the total kinetic energy/total linear momentum/total energy of
the system of two bodies.
Several fundamental laws of mechanics are involved here:
The work done by a conservative force is related to potential energy by
\[ W_{cons} = -\Delta U \]If the work done is positive, then
\[ \Delta U < 0 \]Thus the potential energy decreases.
Correct answer: decreases
Friction is a non-conservative force that converts mechanical energy into thermal energy.
When a body moves against friction, its kinetic energy is reduced.
Correct answer: kinetic energy
For a system of particles,
\[ \frac{d\vec{P}}{dt} = \vec{F}_{ext} \]Internal forces cancel due to Newton's third law.
Correct answer: external force
During an inelastic collision:
Correct answer: total linear momentum
Q7
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the
body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy
of the system.
This question involves important conservation principles in mechanics:
In an elastic collision:
However, the momentum and energy of individual bodies may change during collision.
Statement is False.
Energy of an isolated system remains conserved. However, when external non-conservative forces (such as friction or air resistance) act on a system, mechanical energy may change form.
Thus the statement that energy is always conserved regardless of forces is incorrect in this context.
Statement is False.
For conservative forces such as gravity or spring force:
\[ \oint \vec{F}\cdot d\vec{r} = 0 \]But non-conservative forces such as friction perform non-zero work over closed paths.
Statement is False.
In an inelastic collision:
Hence the final kinetic energy is always less than the initial kinetic energy.
Statement is True.
Q8
Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved
during the short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision
of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance
between their centres, is the collision elastic or inelastic ? (Note, we are talking
here of potential energy corresponding to the force during collision, not gravitational
potential energy).
During a collision, interacting bodies exert very large forces on each other for a short interval of time. Two important conservation laws apply:
However, during the actual contact phase, some kinetic energy may temporarily convert into elastic potential energy due to deformation.
When two billiard balls collide, they undergo slight elastic deformation.
Answer: No. Kinetic energy is not conserved during the contact interval, but it is conserved before and after the collision.
The collision forces between the balls are internal forces forming an action–reaction pair.
\[ \vec F_{12} = -\vec F_{21} \]If external forces are negligible during the short collision time, the total linear momentum remains constant.
Answer: Yes. Total linear momentum is conserved during the collision.
Result:
KE during contact → No
Momentum during contact → Yes
If the interaction potential depends only on the distance between the centres of the balls, the force is a conservative force.
In a conservative interaction, any potential energy stored during compression is fully recovered during separation.
Therefore the collision is elastic.
Q9
A body is initially at rest. It undergoes one-dimensional motion with constant
acceleration. The power delivered to it at time \(t\) is proportional to
(i) \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^2\)
Power is the rate at which work is done. When a force acts on a moving body, instantaneous power depends on the force and velocity.
The body starts from rest.
\[ u = 0 \] For constant acceleration, \[ v = u + at \] \[ v = at \] Force acting on the body: \[ F = ma \] Instantaneous power: \[ P = Fv \] Substituting the values, \[ P = ma \cdot at \] \[ P = ma^2 t \] Thus, \[ P \propto t \]
Q10
A body is moving unidirectionally under the influence of a source of constant power.
Its displacement in time \(t\) is proportional to
(i) \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^2\)
When a constant power source drives motion, the work done increases linearly with time. Since work changes kinetic energy, velocity increases non-linearly.
Q11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force \( \vec{F} \) given by \[ \vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k}\; \text{N} \] where \( \hat{i},\hat{j},\hat{k} \) are unit vectors along the x-, y- and z-axis respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Work done by a force is calculated using the dot product of force and displacement vectors. Only the component of force along the direction of displacement contributes to work.
The force acting on the body is
\[ \vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \]The displacement of the body is along the z-axis.
\[ \vec{s} = 4\hat{k} \] Now calculate work using the dot product: \[ W = \vec{F}\cdot\vec{s} \] \[ W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k}) \] Using unit vector properties: \[ \hat{i}\cdot\hat{k} = 0, \quad \hat{j}\cdot\hat{k} = 0, \quad \hat{k}\cdot\hat{k} = 1 \] Thus, \[ W = -4(\hat{i}\cdot\hat{k}) + 8(\hat{j}\cdot\hat{k}) + 12(\hat{k}\cdot\hat{k}) \] \[ W = 0 + 0 + 12 \] \[ W = 12\; \text{J} \]Q12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = \(9.11\times10^{-31}\) kg, proton mass = \(1.67\times10^{-27}\) kg, \(1\text{ eV} = 1.60\times10^{-19}\) J).
The kinetic energy of a particle moving with velocity \(v\) is given by
Hence, velocity depends on the ratio of kinetic energy to mass. A lighter particle may move faster even with smaller kinetic energy.
Since
\[ v \propto \sqrt{\frac{KE}{m}} \] the velocity ratio can also be written as \[ \frac{v_e}{v_p} = \sqrt{\frac{KE_e}{KE_p}\times\frac{m_p}{m_e}} \] This shortcut avoids calculating velocities separately.Q13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of air) until at half its original height, it attains terminal speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is \(10\,\text{m s}^{-1}\)?
When an object falls through air, two forces act on it:
Work done by gravity depends only on vertical displacement, not on velocity.
Q14 A molecule in a gas container hits a horizontal wall with speed \(200\,\text{m s}^{-1}\) at an angle \(30^\circ\) with the normal and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
When a particle collides with a rigid wall, the velocity component parallel to the wall remains unchanged, while the component normal to the wall reverses direction.
Momentum conservation must be applied to the complete system (molecule + wall).
A pump on the ground floor of a building can pump up water to fill a tank of volume \(30\,\text{m}^3\) in 15 min. If the tank is \(40\,\text{m}\) above the ground and the efficiency of the pump is \(30\%\), how much electric power is consumed by the pump?
When water is pumped to a height, the pump performs work against gravity. The useful work done equals the increase in gravitational potential energy of the water.
Q16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with speed \(V\). If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision?
In an elastic collision, both linear momentum and kinetic energy are conserved. When identical masses collide head-on, energy transfer occurs efficiently between them. This behaviour is observed in devices like a Newton’s cradle.
The momentum and kinetic energy are transmitted through the stationary balls, so the last ball moves forward while the others remain at rest.
Q17 The bob A of a pendulum released from \(30^\circ\) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
When a pendulum bob is released from an angle, gravitational potential energy converts into kinetic energy at the lowest point. If the bob collides elastically with another identical mass, momentum and kinetic energy are conserved.
Q18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is \(1.5\,\text{m}\), what is the speed with which the bob arrives at the lowermost point, given that it dissipates \(5\%\) of its initial energy against air resistance?
When a pendulum bob is released, gravitational potential energy converts into kinetic energy. If air resistance is present, part of the mechanical energy is dissipated.
If there were no air resistance, the speed would be
\[ v = \sqrt{2gL} \approx 5.42\,\text{m/s} \]Air resistance reduces the speed slightly by dissipating mechanical energy.
Q19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while sand starts leaking out of a hole in the floor of the trolley at the rate of \(0.05\,\text{kg s}^{-1}\). What is the speed of the trolley after the entire sand bag is empty?
If mass leaves a system with the same horizontal velocity as the system, no horizontal force or impulse acts on the remaining part. Therefore the velocity of the system remains unchanged.
This situation differs from rocket motion. In rockets, mass is expelled with velocity relative to the rocket, producing thrust. Here the sand simply drops vertically with the same horizontal velocity.
Q20 A body of mass \(0.5\,\text{kg}\) travels in a straight line with velocity \[ v = a x^{3/2} \] where \(a = 5\,\text{m}^{-1/2}\text{s}^{-1}\). What is the work done by the net force during its displacement from \(x=0\) to \(x=2\,\text{m}\)?
The work done by the net force equals the change in kinetic energy of the body. This is the Work–Energy Theorem.
If velocity is given as a function of position, we can directly compute the initial and final speeds and determine the change in kinetic energy.
When velocity is given as a function of position, the Work–Energy theorem provides a quick method to find work without calculating force explicitly.
Q21 The blades of a windmill sweep out a circle of area \(A\). (a) If wind flows with velocity \(v\) perpendicular to the circle, what mass of air passes through it in time \(t\)? (b) What is the kinetic energy of this air? (c) If the windmill converts 25% of this energy into electrical energy and \(A = 30\,\text{m}^2,\ v = 36\,\text{km/h}\), air density \(= 1.2\,\text{kg m}^{-3}\), what electrical power is produced?
Wind possesses kinetic energy because moving air has mass and velocity. When wind passes through the blades of a windmill, a portion of its kinetic energy can be converted into electrical power.
Wind power depends strongly on velocity because
\[ P \propto v^3 \]Thus even a small increase in wind speed greatly increases power generation.
Q22
A person trying to lose weight lifts a 10 kg mass one thousand times to a height of
0.5 m each time. Assume the potential energy lost when lowering the mass is dissipated.
(a) How much work does she do against gravity?
(b) Fat supplies \(3.8\times10^{7}\) J of energy per kilogram which is converted to mechanical
energy with 20% efficiency. How much fat will the dieter use up?
When an object is lifted vertically, work must be done against the gravitational force. The work done equals the increase in gravitational potential energy.
Even a large amount of physical effort corresponds to relatively small fat loss because the human body converts chemical energy into mechanical energy with limited efficiency.
Q23
A family uses \(8\,\text{kW}\) of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of
\(200\,\text{W m}^{-2}\). If \(20\%\) of this energy can be converted into useful electrical energy,
how large an area is needed to supply \(8\,\text{kW}\)?
(b) Compare this area to that of the roof of a typical house.
Solar panels convert a fraction of incident solar radiation into electrical energy. The useful electrical power depends on the incident solar power, panel area, and conversion efficiency.
Solar power generation depends strongly on panel area and efficiency. Improving panel efficiency significantly reduces the required installation area.
This quick revision sheet summarizes all the important formulas from the chapter Work, Energy and Power.
If you remember these 10 results, you can solve most exam problems from this chapter.
This mind map connects all key concepts of the chapter in a single visual structure. Review it before exams for quick conceptual recall.
Most questions in JEE Main, NEET, and other competitive exams from this chapter follow a few standard patterns. Master these patterns to solve exam questions quickly.
| # | Exam Pattern | Key Concept Used |
|---|---|---|
| 1 | Block sliding on rough surface | Work done by friction and Work–Energy theorem |
| 2 | Object falling from height | Conversion of potential energy to kinetic energy |
| 3 | Pendulum speed at lowest point | \(v = \sqrt{2gh}\) using energy conservation |
| 4 | Power required to lift water | \(P = \rho V g h / t\) |
| 5 | Motion under constant power | \(P = Fv\) and velocity–time relation |
| 6 | Elastic collision problems | Momentum + kinetic energy conservation |
| 7 | Inelastic collision problems | Momentum conserved, KE lost |
| 8 | Variable force work calculation | \(W = \int F\,dx\) |
| 9 | Machine efficiency problems | \(\eta = \frac{output}{input}\) |
| 10 | Windmill / solar energy problems | Energy conversion and power formulas |
If you master these ten patterns, you can solve almost every JEE/NEET problem from the Work–Energy–Power chapter.
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