Work and Energy
NCERT Solutions • Class 9 Science • Chapter 10Chapter Exercises
1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
| Activity | Work Done? | Reasoning |
|---|---|---|
| Suma swimming | Yes | Force is applied to push water backward, and Suma is displaced forward. |
| Donkey carrying load | No | The force of gravity (downward) is perpendicular to the displacement (horizontal). Work = 0. |
| Wind-mill lifting water | Yes | Force is applied against gravity to lift (displace) the water. |
| Plant photosynthesis | No | No physical displacement of the plant is taking place. |
| Engine pulling train | Yes | Force is applied by the engine, and the train is displaced in the direction of force. |
| Grains drying in sun | No | No force is causing displacement of grains. |
| Sailboat moving | Yes | Wind energy applies force, causing displacement of the boat. |
2. 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?
Vertical Displacement
Work done by gravity depends only on the vertical displacement ($h$).
Since the initial and final points lie on the same horizontal line, the net vertical displacement is zero ($h = 0$).
Therefore, Work Done $W = mgh = mg(0) =$ Zero.
Work done by gravity depends only on the vertical displacement ($h$).
Since the initial and final points lie on the same horizontal line, the net vertical displacement is zero ($h = 0$).
Therefore, Work Done $W = mgh = mg(0) =$ Zero.
3. A battery lights a bulb. Describe the energy changes involved in the process.
Chemical Energy (in battery) $\rightarrow$ Electrical Energy $\rightarrow$ Heat Energy (in bulb filament) $\rightarrow$ Light Energy.
4. Certain force acting on a 20 kg mass changes its velocity from $5 \text{ m s}^{-1}$ to $2 \text{ m s}^{-1}$. Calculate the work done by the force.
Work-Energy Theorem
Work Done = Change in Kinetic Energy ($K.E.$)
$W = \frac{1}{2}m(v^2 – u^2)$
$W = \frac{1}{2} \times 20 \times (2^2 – 5^2)$
$W = 10 \times (4 – 25) = 10 \times (-21) =$ $-210 \text{ J}$.
(Negative sign indicates force is opposing motion).
Work Done = Change in Kinetic Energy ($K.E.$)
$W = \frac{1}{2}m(v^2 – u^2)$
$W = \frac{1}{2} \times 20 \times (2^2 – 5^2)$
$W = 10 \times (4 – 25) = 10 \times (-21) =$ $-210 \text{ J}$.
(Negative sign indicates force is opposing motion).
5. A mass of 10 kg is at a point 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 = 0 J.
Explanation: The gravitational force acts vertically downwards ($mg$), while the displacement is horizontal. The angle between force and displacement is $90^\circ$.
$W = F s \cos \theta = F s \cos 90^\circ = 0$.
Explanation: The gravitational force acts vertically downwards ($mg$), while the displacement is horizontal. The angle between force and displacement is $90^\circ$.
$W = F s \cos \theta = F s \cos 90^\circ = 0$.
6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
No, it does not violate the law.
As the object falls, its Potential Energy ($PE$) decreases, but its Kinetic Energy ($KE$) increases by an equal amount as its velocity increases. The Total Mechanical Energy ($PE + KE$) remains constant at every point during the fall.
As the object falls, its Potential Energy ($PE$) decreases, but its Kinetic Energy ($KE$) increases by an equal amount as its velocity increases. The Total Mechanical Energy ($PE + KE$) remains constant at every point during the fall.
7. What are the various energy transformations that occur when you are riding a bicycle?
Muscular Energy (of the rider) $\rightarrow$ Mechanical Energy (Kinetic Energy of bicycle) + Heat Energy (due to friction between tyres and road/air resistance).
8. 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?
Work Done is Zero because displacement is zero.
However, Energy Transfer takes place. The muscular energy you spend is converted into:
1. Heat Energy (body heats up).
2. Muscular contraction/fatigue processes internally.
The energy is not transferred to the rock’s motion but dissipated as heat.
However, Energy Transfer takes place. The muscular energy you spend is converted into:
1. Heat Energy (body heats up).
2. Muscular contraction/fatigue processes internally.
The energy is not transferred to the rock’s motion but dissipated as heat.
9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
1 Unit = 1 kWh = $3.6 \times 10^6 \text{ J}$.
250 Units = $250 \times 3.6 \times 10^6 \text{ J}$.
$= 900 \times 10^6 \text{ J} =$ $9 \times 10^8 \text{ J}$.
250 Units = $250 \times 3.6 \times 10^6 \text{ J}$.
$= 900 \times 10^6 \text{ J} =$ $9 \times 10^8 \text{ J}$.
10. An object of mass 40 kg is raised to a height of 5 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.
Potential Energy (At 5m)
$PE = mgh = 40 \times 10 \times 5 =$ $2000 \text{ J}$ (Taking $g=10 \text{ m/s}^2$).
Kinetic Energy (Half-way)
At half-way ($h=2.5 \text{ m}$), Potential Energy is halved: $PE_{half} = 1000 \text{ J}$.
By Conservation of Energy, Loss in PE = Gain in KE.
Therefore, KE = 1000 J.
$PE = mgh = 40 \times 10 \times 5 =$ $2000 \text{ J}$ (Taking $g=10 \text{ m/s}^2$).
Kinetic Energy (Half-way)
At half-way ($h=2.5 \text{ m}$), Potential Energy is halved: $PE_{half} = 1000 \text{ J}$.
By Conservation of Energy, Loss in PE = Gain in KE.
Therefore, KE = 1000 J.
11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Work Done is Zero.
Justification: The gravitational force acts towards the center of the Earth (centripetal force), while the displacement of the satellite at any instant is tangential to the circular path. The angle between Force ($F$) and Displacement ($s$) is $90^\circ$.
$W = F s \cos 90^\circ = 0$.
Justification: The gravitational force acts towards the center of the Earth (centripetal force), while the displacement of the satellite at any instant is tangential to the circular path. The angle between Force ($F$) and Displacement ($s$) is $90^\circ$.
$W = F s \cos 90^\circ = 0$.
12. Can there be displacement of an object in the absence of any force acting on it? Think.
Yes.
If an object is moving with uniform velocity in a straight line, it has displacement, but the net force acting on it is zero ($F = ma$, since $a=0$, $F=0$). So, an object can continue to move (displace) without a continuous unbalanced force.
If an object is moving with uniform velocity in a straight line, it has displacement, but the net force acting on it is zero ($F = ma$, since $a=0$, $F=0$). So, an object can continue to move (displace) without a continuous unbalanced force.
13. 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.
No Work Done.
Justification: Work is defined as $W = F \times s$. Although the person applies an upward force to hold the bundle, there is no displacement of the bundle ($s=0$). Therefore, $W = 0$. The tiredness is due to physiological energy consumption, not mechanical work on the bundle.
Justification: Work is defined as $W = F \times s$. Although the person applies an upward force to hold the bundle, there is no displacement of the bundle ($s=0$). Therefore, $W = 0$. The tiredness is due to physiological energy consumption, not mechanical work on the bundle.
14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Power ($P$) = 1500 W = 1.5 kW.
Time ($t$) = 10 h.
Energy ($E$) = $P \times t = 1.5 \text{ kW} \times 10 \text{ h} =$ $15 \text{ kWh}$ (or 15 Units).
Time ($t$) = 10 h.
Energy ($E$) = $P \times t = 1.5 \text{ kW} \times 10 \text{ h} =$ $15 \text{ kWh}$ (or 15 Units).
15. Illustrate the law of conservation of energy… when we draw a pendulum bob to one side and allow it to oscillate. Why does it eventually stop? Violation?
1. Energy Changes: At extreme positions, Energy is purely Potential ($PE$). At the mean position, Energy is purely Kinetic ($KE$). In between, it is a mix of both. Total Energy remains constant.
2. Why it stops: It stops due to air resistance and friction at the pivot point.
3. Fate of Energy: The mechanical energy is dissipated as heat energy into the surroundings.
4. Violation?: No, it is not a violation. Energy is not lost, just transformed into heat.
2. Why it stops: It stops due to air resistance and friction at the pivot point.
3. Fate of Energy: The mechanical energy is dissipated as heat energy into the surroundings.
4. Violation?: No, it is not a violation. Energy is not lost, just transformed into heat.
16. 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?
According to Work-Energy Theorem:
Work Done = Change in Kinetic Energy
$W = \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2$
$W = 0 – \frac{1}{2}mv^2$
Work to be done (magnitude) = $\frac{1}{2}mv^2$.
Work Done = Change in Kinetic Energy
$W = \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2$
$W = 0 – \frac{1}{2}mv^2$
Work to be done (magnitude) = $\frac{1}{2}mv^2$.
17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Mass ($m$) = 1500 kg.
Velocity ($v$) = $60 \text{ km/h} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s}$.
Work = Kinetic Energy to be removed = $\frac{1}{2}mv^2$
$W = \frac{1}{2} \times 1500 \times (\frac{50}{3})^2$
$W = 750 \times \frac{2500}{9} = \frac{1875000}{9} \approx$ $208333.3 \text{ J}$.
Velocity ($v$) = $60 \text{ km/h} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s}$.
Work = Kinetic Energy to be removed = $\frac{1}{2}mv^2$
$W = \frac{1}{2} \times 1500 \times (\frac{50}{3})^2$
$W = 750 \times \frac{2500}{9} = \frac{1875000}{9} \approx$ $208333.3 \text{ J}$.
18. In each of the following a force F is acting on an object… state whether work done is negative, positive or zero.
Assuming standard diagrams:
1. Force $\perp$ Displacement: Zero Work.
2. Force in direction of Displacement: Positive Work.
3. Force opposite to Displacement: Negative Work.
1. Force $\perp$ Displacement: Zero Work.
2. Force in direction of Displacement: Positive Work.
3. Force opposite to Displacement: Negative Work.
19. 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?
Yes, I agree.
If several forces act on an object but their resultant (net) force is zero (balanced forces), then according to Newton’s Second Law ($F_{net} = ma$), the acceleration will be zero.
If several forces act on an object but their resultant (net) force is zero (balanced forces), then according to Newton’s Second Law ($F_{net} = ma$), the acceleration will be zero.
20. Find the energy in joules consumed in 10 hours by four devices of power 500 W each.
Total Power = $4 \times 500 \text{ W} = 2000 \text{ W} = 2 \text{ kW}$.
Time = 10 h.
Energy (in kWh) = $2 \text{ kW} \times 10 \text{ h} = 20 \text{ kWh}$.
Energy (in Joules) = $20 \times 3.6 \times 10^6 \text{ J}$
$=$ $7.2 \times 10^7 \text{ J}$.
Time = 10 h.
Energy (in kWh) = $2 \text{ kW} \times 10 \text{ h} = 20 \text{ kWh}$.
Energy (in Joules) = $20 \times 3.6 \times 10^6 \text{ J}$
$=$ $7.2 \times 10^7 \text{ J}$.
21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
When the object hits the ground, its Kinetic Energy is converted into:
1. Heat Energy (ground and object get warm).
2. Sound Energy (thud sound).
3. Potential Energy of deformation (if the object or ground deforms).
1. Heat Energy (ground and object get warm).
2. Sound Energy (thud sound).
3. Potential Energy of deformation (if the object or ground deforms).