Work, Energy, and Power
Work
Definition: Work is performed when a force acts on an object and causes it to move a certain distance.
Mathematical Expression: Work done, W = F × s × cosθ
F: Force applied
s: Displacement of the object
θ = Angle between the displacement vectors & force
Unit: Joule (J)
Energy
Definition: Energy is the potential to exert force and cause motion.
Types of Energy:
Kinetic Energy:The energy of motion.
Formula: KE = 1/2 mv²
Gravitational Potential Energy: PE = mgh
Elastic Potential Energy: Energy stored in a stretched or compressed elastic object.
Definition: Power measures how quickly work is done or energy is used.
Mathematical Expression: Power, P = W/t
W: Work done
t: Time taken
Unit: Watt (W)
Key Points:
Work, energy, and power are scalar quantities.
The net work done on an object causes a change in its kinetic energy, as described by the work-energy theorem.
The principle of conservation of mechanical energy states that the total mechanical energy (kinetic + potential) of a system remains constant in the absence of non-conservative forces.
Understanding these concepts is essential for analyzing various physical phenomena, from the motion of celestial bodies to the operation of everyday machines.
1. 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.
Ans : (a) Positive:The man applies a force upwards, and the bucket moves upwards. The angle between the force and displacement vectors is 0 degrees, so the cosine of the angle is 1. Therefore, the work done is positive.
(b)Negative: The gravitational force acts downwards, while the displacement of the bucket is upwards.
(c)Negative: Friction opposes the motion of the body, As friction acts opposite to the direction of motion, the work done by friction is negative.
(d) Positive: The applied force acts in the direction of motion of the body to overcome friction and maintain uniform velocity.
(e) Negative: The resistive force of air acts opposite to the direction of motion of the pendulum. This force gradually reduces the pendulum’s kinetic energy, bringing it to rest. Hence, the resistive force does negative work.
2. 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.
Ans : (a) **Calculating Work Done by Applied Force:**
Friction:** Frictional force = μN = 0.1 * 2 kg * 9.8 m/s² = 1.96 N
Net Force:** Net force = Applied force – Frictional force = 7 N – 1.96 N = 5.04 N
Acceleration:** Using Newton’s second law, a = F_net / m = 5.04 N / 2 kg = 2.52 m/s²
Distance:** Using the equation of motion, s = ut + 0.5at², we get s = 0.5 * 2.52 * 10² = 126 m
*Work done:** W = F * s = 7 N * 126 m = 882 J
(b)Work Done by Friction:**
* W_friction = frictional force * distance = -1.96 N * 126 m = -246.96 J (Negative sign indicates work done against the motion)
(c) Work Done by Net Force:**
* W_net = net force * distance = 5.04 N * 126 m = 635.04 J
(d)Change in Kinetic Energy: * The change in kinetic energy equals the work done by the net force, which is 635.04 J.
3 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
Ans : Analyzing the Potential Energy Functions
Figure 5.11(a):
Regions where the particle cannot be found: None.
Minimum total energy: V₀
Physical context: A particle in a potential well. The particle can exist anywhere in the well, with a minimum energy equal to the potential energy at the bottom of the well.
Figure 5.11(b):
Regions where the particle cannot be found: x < a and x > d
Minimum total energy: V₁
Physical context: A particle in a potential well with finite walls. The particle is confined to the region between a and d. The minimum energy required to escape the well is V₁.
Figure 5.11(c):
Regions where the particle cannot be found: x < 0 and x > b
Minimum total energy: V₀
Physical context: A particle in a potential well with infinite walls. The particle is strictly confined to the region between 0 and b. The minimum energy is the ground state energy of the well.
Figure 5.11(d):
Regions where the particle cannot be found: x < -x₁ and x > x₂
Minimum total energy: V₁
Physical context: A particle in a double-well potential. The particle can be trapped in either of the two wells, with a minimum energy of V₁ to escape.
In each case, the particle cannot exist in regions where the potential energy exceeds its total energy, as this would imply negative kinetic energy, which is physically impossible.
4 .The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1 , the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Ans : Given:
* Force constant, k = 0.5 Nm⁻¹
* Total energy, E = 1 J
given,
potential energy of particle
V(x) = (1/2)kx²
When the particle reaches its maximum displacement (x = ±xm), its total energy is converted into potential energy. Therefore, at the turning points:
Total energy = Potential energy
1 J = (1/2)(0.5 Nm⁻¹)(xm)²
Solving for xm:
xm² = 4
xm = ± 2 m
Hence, the particle will turn back when it reaches x = ±2 m.
5 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 ?
Ans : (a) The heat energy required for burning the rocket casing is obtained at the expense of the rocket’s kinetic energy. As the rocket burns fuel, it loses mass, and this loss in mass reduces its kinetic energy. This kinetic energy is then converted into heat energy to burn the casing.
(b) Gravitational force is a conservative force. This means that the work done by a conservative force in moving a particle between two points is independent of the path taken. In the case of a comet orbiting the sun, the gravitational force is always directed towards the sun, and its work done over a complete orbit is zero, as the net displacement of the comet is zero.
(c) As the satellite loses energy due to atmospheric resistance, it moves closer to the Earth. This decrease in distance from the Earth increases the gravitational potential energy of the satellite. However, the total mechanical energy (kinetic + potential) of the satellite decreases. To conserve this total mechanical energy, the kinetic energy of the satellite must increase, resulting in an increase in its speed.
(d) In Fig. 5.13(i), the man is doing work against gravity to lift the 15 kg mass. In Fig. 5.13(ii), the man is doing work against gravity to lift the 15 kg mass, but he is also gaining work from the pulley system, which reduces the effort required. Therefore, the work done in case (i) is greater than the work done in case (ii).
6 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.
Ans : (a) A conservative force doing positive work on a body leads to a decrease in its potential energy.
(b)Work done against friction always results in a loss of kinetic energy.
(c) The rate of change of a system’s total momentum is directly proportional to the net external force acting on the system.
(d) In an inelastic collision, the total linear momentum of the system remains conserved.
7 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.
Ans : (a) False. In an elastic collision, the total kinetic energy and the total momentum of the system are conserved, not the individual kinetic energy and momentum of each body.
(b) False. Total energy is conserved only when there are no non-conservative forces acting on the system. Internal forces like electrostatic or gravitational forces conserve energy, but external forces like friction or air resistance can dissipate energy.
(c) False. For non-conservative forces like friction, the work done over a closed loop is not zero.
(d) True.During an inelastic collision, a portion of the initial kinetic energy is transformed into other energy forms like heat, sound, or deformation, resulting in a decrease in the total kinetic energy of the system.
8. 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 ? Fig. 5.13 Fig. 5.12 2024-25 90 PHYSICS
(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).
Ans : (a) During the brief moment of collision, the total kinetic energy is not conserved. A portion of this kinetic energy is temporarily transformed into potential energy as the balls deform. However, this potential energy is subsequently reconverted into kinetic energy, leading to the conservation of total kinetic energy after the collision.
(b) The total linear momentum of the system remains conserved during the brief duration of an elastic collision.This is because there are no external forces acting on the system of the two balls during the collision.
(c) For an inelastic collision:
– Total kinetic energy is not conserved. Some of the kinetic energy is lost to other forms of energy like heat, sound, or deformation of the colliding objects.
-The total linear momentum of the system remains constant as no external forces are acting on it.
(d) If the potential energy depends only on the separation distance between the centers of the balls, it implies that the force between them is conservative. In such a case, the total mechanical energy (kinetic + potential) is conserved, and the collision is elastic.
9 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
Ans : We know that power, P = F × v
where,
F = force
v = velocity.
Using Newton’s second law, F = ma
where,
m = mass
a = acceleration.
Given that the acceleration is constant, let’s denote it as ‘a’.
Therefore, P = ma × v
Now, we can relate velocity to acceleration and time using the equation of motion:
v = u + at
Since the body starts from rest, u = 0.
So, v = at
Substituting this into the power equation:
P = ma × at = ma²t
Since m and a are constants, P ∝ t.
Therefore, the power delivered to the body is proportional to t.
Hence, the correct option is (ii) t.
10. 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
Ans : We know that power, P = work done/time = (force × displacement)/time
For constant power, P = constant.
Let’s assume the force acting on the body is constant (F).
So, F × displacement/time = constant
Or, displacement ∝ time
Therefore, the displacement is proportional to t.
Hence, the correct option is (ii) t.
11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
where k ,j ,i ˆ ˆ ˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
Ans : Since the body is displaced 4 m along the z-axis only, the displacement vector is given by:
S = 0i + 0j + 4k
Given,
force acting on the body
F = -i + 2j + 3k
find work done by the force by using the dot product:
W = F · S = (-i + 2j + 3k) · (0i + 0j + 4k)
= 12 Joules
Hence, the work done by the force is 12 Joules.
12. 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 = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).
Ans : Here, the kinetic energy of the electron (Ke) is 10 keV and the kinetic energy of the proton (Kp) is 100 keV. The mass of the electron (me) is 9.11 × 10^-31 kg and the mass of the proton (mp) is 1.67 × 10^-27 kg.
We know that kinetic energy (K) is related to mass (m) and velocity (v) by the formula:
K = (1/2)mv²
Rearranging the formula to solve for velocity:
v = √(2K/m)
Therefore, the ratio of the velocities of the electron (ve) and the proton (vp) is:
ve/vp = √(Ke/Kp) × √(mp/me)
Substituting the given values:
ve/vp = √(10 keV / 100 keV) × √(1.67 × 10^-27 kg / 9.11 × 10^-31 kg)
≈ 13.54
This means that the electron is traveling approximately 13.54 times faster than the proton.
13. 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 the air) until at half its original height, it attains its maximum (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 m s–1 ?
Ans : Given:
raindrop Radius is r = 2 mm
= 2 × 10⁻³ m
Distance traveled in each half of the journey,
S = 500 m / 2 = 250 m
water Density, ρ = 10³ kg/m³
Calculating the mass of the raindrop:**
Volume of the raindrop is (4/3)πr³
= (4/3)π(2 × 10⁻³)³ m³
Mass of the raindrop, m = Volume × Density
= (4/3)π(2 × 10⁻³)³ × 10³ kg ≈ 3.35 × 10⁻⁵ kg
Work done by gravity in each half:Work done by gravity, W = mgh = (3.35 × 10⁻⁵ kg) × (9.8 m/s²) × (250 m) ≈ 0.082 J
Note:** The work done by gravity remains the same regardless of whether the raindrop falls with decreasing acceleration or uniform speed.
Work done by resistive forces:If there were no resistive forces, the energy of the raindrop on reaching the ground would be:
E₁ = mgh = (3.35 × 10⁻⁵ kg) × (9.8 m/s²) × (500 m) ≈ 0.164 J
Actual energy of the raindrop on reaching the ground (considering resistive forces):
E₂ = (1/2)mv² = (1/2)(3.35 × 10⁻⁵ kg)(10 m/s)²
≈ 1.675 × 10⁻³ J
Work done by resistive forces, W_resistive = E₁ – E₂ = 0.164 J – 1.675 × 10⁻³ J
≈ 0.1623 J
14. A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Ans : Let’s denote the mass of the molecule as m and the mass of the wall as M. As the wall is significantly more massive, it remains stationary.
Resolving the molecule’s momentum into its x and y components, we get:
15. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?
Ans : Understanding the Problem:Volume of the tank = 30 m³
Time taken to fill the tank = 15 min = 900 s
Height of the tank = 40 m
Efficiency of the pump = 30%
*Solution:
1. *Mass of water pumped Density of water = 1000 kg/m³
water Mass is calculated by using formula Density × Volume
= 30 m³ × 1000 kg/m³
= 30,000 kg
2.Work done in pumping water:
Work done = mgh = 30,000 kg × 9.8 m/s² × 40 m = 11,760,000 J
3. *Power required to pump water
* Power = Work done / Time = 11,760,000 J / 900 s ≈ 13,067 W
4. Electric power consumed:
* Since the pump is 30% efficient, the actual power consumed will be:
Electric power = Power required / Efficiency = 13,067 W / 0.3
≈ 43,557 W
Therefore, the electric power consumed by the pump is approximately 43,557 Watts.
16. 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 a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ?
Ans : m represent the mass of each ball bearing.
Before collision:
Total kinetic energy (KE) = (1/2)mv² + 0 = (1/2)mv²
After collision
Case I:**
KE₁ = (1/2)(2m)(v/2)²
= (1/4)mv²
Case II
KE₂ = (1/2)mv²
*Case III:
* KE₃ = (1/2)(3m)(v/3)²
= (1/6)mv²
Analysis:
Since kinetic energy is conserved in an elastic collision, only Case II satisfies this condition. In Case I and Case III, the final kinetic energy is less than the initial kinetic energy, indicating that energy is lost, which is not possible in an elastic collision.
Therefore, only Case II is a possible outcome of the elastic collision.
17. The bob A of a pendulum released from 30o 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 the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.
Ans : The elastic collision causes bob A to come to rest while imparting its entire velocity to bob B.
Conservation of momentum: The initial momentum of bob A is transferred entirely to bob B.
*Conservation of kinetic energy. As bob A comes to rest, all its kinetic energy is transferred to bob B.
Therefore, bob A will not rise again after the collision.
18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?
Ans : *Understanding the Problem:**
We have a pendulum with a length of 1.5 m. The bob is released from a horizontal position, meaning it starts at its highest point. We need to find its speed at the lowest point, considering a 5% energy loss due to air resistance.
Solution:
1. Initial Potential Energy:
When the bob is at its highest point, it has maximum potential energy.
Potential Energy (PE) = mgh
where m is the mass of the bob, g is the acceleration due to gravity (9.8 m/s²), and h is the height (equal to the length of the pendulum in this case).
2. *Final Kinetic Energy:
At the lowest point, all the potential energy (minus the 5% loss) is converted into kinetic energy.
Kinetic Energy (KE) = (1/2)mv²
v = final velocity of the bob.
3.Energy Conservation (with 5% loss) 0.95 * Initial PE = Final KE
0.95 * mgh = (1/2)mv²
4. Solving for v (m) mass = cancels out on both sides
0.95 * gh = (1/2)v²
v = √(2 * 0.95 * gh)
Substituting the values:
v = √(2 * 0.95 * 9.8 m/s² * 1.5 m)
≈ 5.22 m/s
Therefore, the speed of the bob at the lowest point is approximately 5.22 m/s.
19. 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 on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ?
Ans : The system comprising the trolley and the sandbag is moving with a constant velocity. This indicates that there is no net external force acting on the system. As the sand leaks out, it does not exert any external force on the system. Hence, the total momentum of the system is conserved.. Since the mass of the system decreases while the momentum remains constant, the velocity of the trolley must increase to compensate for the mass loss.
20. A body of mass 0.5 kg travels in a straight line with velocity v =a x 3/2 where a = 5 m–1/2 s–1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?
Ans : Here, the mass of the body, m = 0.5 kg, and the acceleration,
a = 5 m^(-1/2)s^(-1).
The velocity of the body is given as a function of displacement: v = ax^(3/2)
Initial velocity (at x = 0)
v₁ = a(0)^(3/2)
= 0
**Final velocity (at x = 2 m):**
v₂ = a(2)^(3/2) = 5 × 2^(3/2) m/s
Work done:
Since the work done is equal to the change in kinetic energy:
Work done = (1/2)m(v₂² – v₁²)
= (1/2) × 0.5 kg × [(5 × 2^(3/2))² – 0]
≈ 50 J
Therefore, the work done by the net force is approximately 50 Joules.
21. The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2 , v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?
Ans : (a) Av stands for Volume of wind flowing (per second )
Mass of wind flowing per second = ρAv
(b) Kinetic energy of air:
KE = (1/2)mv² = (1/2)(ρAv)v²
= (1/2)ρAv³
(c) Electrical power produced:
Power = 25% of kinetic energy per second
= 0.25 × (1/2)ρAv³
= (1/8)ρAv³
Substituting the given values:
A = 30 m², v = 36 km/h = 10 m/s, ρ = 1.2 kg/m³
Electrical power = (1/8) × 1.2 kg/m³ × 30 m² × (10 m/s)³ = 4500 W
= 4.5 kW
22. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate.
How much fat will the dieter use up?
Ans : (a) Work done against gravity in one lift:
W₁ = mgh = 10 kg × 9.8 m/s² × 0.5 m = 49 J
Total work done in 1000 lifts:
W_total = 1000 × W₁
= 49000 J
(b) Energy supplied by 1 kg of fat:
E_fat = 3.8 × 10⁷ J/kg × 20% = 0.76 × 10⁷ J/kg
Mass of fat used:
Mass = Total work done / Energy per kg of fat
= 49000 J / 0.76 × 10⁷ J/kg
≈ 6.45 × 10⁻³ kg
23. A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
Ans : (a) Calculating the required area:
* Power required = 8 kW = 8000 W
* Usable solar power per square meter = 20% of 200 W/m² = 40 W/m²
* Required area = Power required / Usable power per square meter = 8000 W / 40 W/m² = 200 m²
(b) Comparing with a typical roof area:
A typical house roof area is much smaller than 200 square meters. This indicates that solar power systems, while environmentally friendly, require significant space to generate substantial amounts of electricity. Additional factors like weather conditions, geographical location, and panel efficiency also influence the actual power output of a solar system.
