Q1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term “work.”
(a) Suma is swimming in a pond.
(b) A donkey is carrying a load on its back.
(c) A windmill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
(e) An engine is pulling a train.
(f) Food grains are getting dried in the sun.
(g) A sailboat is moving due to wind energy.

Answer:
(a) Swimming: No work on water by the swimmer—each stroke pushes water back, but water is displaced. However, work done by swimmer on water.
(b) Donkey carrying load: No work on load in vertical direction if load moves horizontally. Work done on donkey’s muscles but not on load.
(c) Windmill lifting water: Yes work = weight × height.
(d) Photosynthesis: No mechanical work, but chemical energy storage—outside mechanical definition.
(e) Engine pulling train: Yes, force × displacement along track.
(f) Drying grains: No mechanical work; energy transfer as heat, but not mechanical work.
(g) Sailboat moving: Yes, wind force on sails causes displacement; work done by wind on boat.

Q2. 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?

Answer: Zero. Gravity is a conservative force; net work over any horizontal displacement returns to zero potential change.

Q3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer: Chemical energy in the battery is converted into electrical energy in the circuit, which is then converted into light and heat energy in the bulb filament.

Q4. Certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. Calculate the work done by the force.
Answer:

Given:
Mass,

$m = 20$

kg
Initial velocity,

$u = 5$

m/s
Final velocity,

$v = 2$

m/s
Work done,

$W = ?$

Formula:

$$W = \frac{1}{2} m (v^2 – u^2)$$

Substitution:

$$W=\frac{1}{2}\times 20\times (2^2-5^2)$$

$$=10\times (4-25)$$

$$=10\times (-21)$$

$$W = \frac{1}{2} \times 20 \times (2^2 – 5^2) = 10 \times (4 – 25) = 10 \times (-21) = -210 \text{ joules}$$

Work done by the force = -210 joules (negative sign indicates the force reduced the object’s kinetic energy).

Q5. 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.

Answer: Zero. Gravitational force is vertical; displacement is horizontal, so no component of displacement in direction of force.

Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer: No. The decrease in potential energy converts into an equal increase in kinetic energy, keeping total mechanical energy constant (neglecting air resistance).

Q7. What are the various energy transformations that occur when you are riding a bicycle?

Answer: Chemical energy in muscles → mechanical energy of pedalling → kinetic energy of bicycle → some energy into heat due to friction in chain and air resistance.

Q8. 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?

Answer: Muscular chemical energy is expended, but no mechanical work on the rock. Energy is converted into heat in muscles and dissipated in the body; no energy transferred to rock.

Q9. A household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer: 1 unit = 1 kWh = 3.6 × 10⁶ J
250 units = 250 × 3.6 × 10⁶ J = 9.0 × 10⁸ J.

Q10. 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.

Answer:
Given: m = 40 kg, h = 5 m, g = 9.8 m s⁻²
PE = m g h = 40 × 9.8 × 5 = 1960 J
At halfway (h = 2.5 m), potential energy = 40 × 9.8 × 2.5 = 980 J
Total mechanical energy = 1960 J, so KE at halfway = 1960 – 980 = 980 J.

Q11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer: Zero, because the gravitational force is always perpendicular to the instantaneous displacement in a circular orbit, so no work is done.

Q12. 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.

Answer: Yes. An object can move with constant velocity in the absence of net force (Newton’s first law). Displacement occurs but no work is done.

Q13. 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.

Answer: No work done on the hay because there is no displacement. The person expends biological energy converting to heat but no mechanical work on the hay.

Q14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer: Power = 1500 W = 1.5 kW, time = 10 h
Energy = Power × time = 1.5 kW × 10 h = 15 kWh = 15 units.

Q15. 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?

Answer: At maximum displacement, energy is all potential. As it swings down, potential converts to kinetic. At lowest point, all kinetic. Then converts back to potential at opposite amplitude. The bob eventually stops due to non-conservative forces (air resistance and friction at pivot) converting mechanical energy to heat and sound. Total energy remains conserved when accounting for these forms, so no violation occurs.

Q16. 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?

Answer: Initial kinetic energy = ½ m v². To bring to rest, work done by opposing force = –½ m v². Magnitude of work required = ½ m v².

Q17. Calculate the work required to stop a car of 1500 kg moving at a velocity of 60 km/h.
Answer:

Given:
Mass of car,

$m = 1500$

kg
Initial velocity,

$u = 60$

km/h =

$\frac{60 \times 1000}{3600} = 16.67$

m/s
Final velocity,

$v = 0$

m/s (car stops)
Work done,

$W = ?$

Formula:
Work done to stop the car equals the change in kinetic energy:

$$W = \frac{1}{2} m (v^2 – u^2)$$

Substitution:

$$W=\frac{1}{2}\times 1500\times (0^2-{16.67}^2)$$

$$=750\times (-277.89)$$

$$W = \frac{1}{2} \times 1500 \times (0^2 – 16.67^2) = 750 \times (-277.89) = -208,417.5 \text{ joules}$$

Work required to stop the car = 208,417.5 joules (the negative sign indicates work done against the motion to bring the car to rest).

Q18. In each of the following a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
(a) Force F pointing east.
(b) Force F pointing north.
(c) Force F pointing west.

Answer:
(a) Positive work (force and displacement same direction).
(b) Zero work (force perpendicular to displacement).
(c) Negative work (force opposite displacement).

Q19. Soni says that the acceleration in an object could be zero when several forces are acting on it. Do you agree with her? Why?

Answer: Yes. If the vector sum of all forces is zero, net force is zero and acceleration = 0 (Newton’s first law). The object moves with constant velocity or remains at rest.

Q20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.

Answer: Each device power = 500 W = 0.5 kW, time = 10 h. Energy per device = 0.5 × 10 = 5 kWh. For four devices, total = 4 × 5 = 20 kWh.

Q21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer: Its kinetic energy is converted into other forms—heat, sound, deformation of ground or object—so total energy is conserved, fulfilling the conservation of energy.