Numericals with Answers - Work and Energy | Science Class 9 PDF Download

The study of Class 9 Science  Chapter - Work and Energy  explores how forces cause motion and how energy transforms in the world around us. You'll learn about the scientific meaning of work, the different forms of energy like kinetic and potential energy, the law of conservation of energy, and the concept of power. 

These ideas will help you understand everyday phenomena-like lifting objects, moving vehicles, or even a swinging Pendulum-and how energy is transferred and conserved. 

Basic Formulae

Work Done: Work is done when a force causes displacement in its direction.

• If force and displacement are in the same direction: W = F × S
• If force is at an angle θ to the displacement: W = F × S × cos(θ)
• Unit: Joule (J), where 1 J = 1 Nm.
• Work is zero if displacement is zero or if the force is Perpendicular to the displacement (cos 90° = 0).
• Work is positive if force and displacement are in the same direction, negative if opposite.

Kinetic Energy (K.E.): The energy of an object due to its motion.
K.E. = ½mv²
m = mass (kg), v = velocity (m/s), K.E. in Joules (J).

Potential Energy (P.E.): The energy of an object due to its position (height).
P.E. = mgh
m = mass (kg), g = acceleration due to gravity (use 10 m/s² unless specified), h = height (m), P.E. in Joules (J).

Law of Conservation of Energy: Energy cannot be created or destroyed; it transforms from one form to another.
Total mechanical energy (K.E. + P.E.) remains constant in a system with no external forces.
K.E. + P.E. = constant.

Power: The rate of doing work or transferring energy.
P = W / t
W = work done (J), t = time (s), P in Watts (W), 1 W = 1 J/s.
Larger units: 1 kW = 1000 W, 1 MW = 10⁶ W, 1 hp = 746 W.

Solved Examples

Q1: An object of mass 10 kg is moving at a velocity of 20 m/s. If the object comes to rest after covering 40 m, calculate the average retarding force.
Sol: Initial speed, v = 20 m/s; mass, m = 10 kg; displacement, s = 40 m.
Initial kinetic energy, K.E. = ½ m v².
K.E. = ½ × 10 × (20)² = 0.5 × 10 × 400 = 5 × 400 = 2000 J.
Final kinetic energy = 0 (object comes to rest).
Change in kinetic energy = 0 - 2000 = -2000 J. This equals the work done by the retarding force.
Work by retarding force = -F × s = -2000 J.
⇒ -F × 40 = -2000.
⇒ F = 2000 / 40 = 50 N.
Average retarding force = 50 N (opposite to motion).

Q2: A boy pulls a cart with a force of 30 N at an angle of 60° to the horizontal. If the cart moves 8 m horizontally, calculate the work done by the boy. (Use cos 60° = 0.5)
Sol: Formula: W = F × S × cos(θ).
Given: F = 30 N, S = 8 m, cos 60° = 0.5.
First compute F × S = 30 × 8 = 240.
W = 240 × 0.5 = 120 J.
Work done by the boy = 120 J.

Q3: A crane does 18000 J of work in 15 seconds. Calculate its power in kilowatts (kW).
Sol: Formula: P = W / t.
Given: W = 18000 J, t = 15 s.
P = 18000 / 15 = 1200 W.
Convert to kilowatts: 1 kW = 1000 W.
P = 1200 / 1000 = 1.2 kW.

Q4: A stone of mass 2 kg is dropped from a height of 8 m. Calculate the velocity of the stone just before it hits the ground, assuming no air resistance. (Take g = 10 m/s²)

Sol: Use conservation of mechanical energy: P.E. at top = K.E. just before hitting the ground.
P.E. = mgh = 2 × 10 × 8 = 160 J.
Let v be the speed just before impact. Then K.E. = ½ m v² = 160 J.
½ × 2 × v² = 160 ⇒ v² = 160.
v = √160 = √(16 × 10) = 4√10 ≈ 4 × 3.16 = 12.64 m/s.
Velocity just before impact ≈ 12.64 m/s.

Q5: A Pendulum bob of mass 0.5 kg is released from a height of 4 m above its lowest point. Calculate its potential energy at the top, kinetic energy at the lowest point, and kinetic energy when it is 1 m above the lowest point. (Take g = 10 m/s²)
Sol: Step 1: Potential energy at the top.
m = 0.5 kg, g = 10 m/s², h = 4 m.
P.E. = m g h = 0.5 × 10 × 4 = 20 J.

Step 2: Kinetic energy at the lowest point (h = 0).
Total mechanical energy at top = P.E. + K.E. = 20 + 0 = 20 J.
At lowest point, P.E. = 0, so K.E. = total energy = 20 J.

Step 3: Kinetic energy at h = 1 m above the lowest point.
P.E. at 1 m = m g h = 0.5 × 10 × 1 = 5 J.
Total energy = 20 J, so K.E. = 20 - 5 = 15 J.
Answers: P.E. at top = 20 J; K.E. at lowest point = 20 J; K.E. at 1 m above lowest point = 15 J.

Q6: A lift raises 1500 kg of material to a height of 12 m in 1 minute. Calculate the power of the lift in watts. (Take g = 10 m/s²)
Sol: Step 1: Calculate work done (work = potential energy gained).
W = m g h = 1500 × 10 × 12 = 180000 J.
Step 2: Convert time to seconds.
t = 1 minute = 60 s.
Step 3: Calculate power.
P = W / t = 180000 / 60 = 3000 W.
Power of the lift = 3000 W.

Q7: An object at a height of 4 m above the ground has a potential energy of 800 J. Calculate the mass of the object. (Take g = 10 m/s²)
Sol: Use P.E. = m g h.
Given: P.E. = 800 J, g = 10 m/s², h = 4 m.
800 = m × 10 × 4 = 40 m.
m = 800 / 40 = 20 kg.
Mass = 20 kg.

Q8: Two objects, A and B, have masses 3 kg and 6 kg, respectively, and are moving with velocities 5 m/s and 3 m/s. Compare their kinetic energies.
Sol: Formula: K.E. = ½ m v².
For object A: m = 3 kg, v = 5 m/s.
K.E._A = ½ × 3 × (5)² = 0.5 × 3 × 25 = 1.5 × 25 = 37.5 J.
For object B: m = 6 kg, v = 3 m/s.
K.E._B = ½ × 6 × (3)² = 0.5 × 6 × 9 = 3 × 9 = 27 J.
Ratio K.E._A : K.E._B = 37.5 : 27 ≈ 1.39 : 1.
Object A has greater kinetic energy (≈ 1.39 times that of B).
K.E. of A : K.E. of B ≈ 1.39 : 1

Q9: A sled is pushed with a force of 70 N at an angle of 30° to the horizontal for 20 m forward, then pushed back 10 m with the same force and angle. Calculate the total work done by the force. (Use cos 30° = 0.866, cos 150° = -0.866)
Sol: Forward motion:
F = 70 N, S = 20 m, cos 30° = 0.866.
W₁ = F × S × cos 30° = 70 × 20 × 0.866 = 1400 × 0.866 = 1212.4 J.
Backward motion:
Here the displacement is opposite the original direction, so the angle between force and displacement is 150° and cos 150° = -0.866.
S = 10 m.
W₂ = 70 × 10 × (-0.866) = 700 × (-0.866) = -606.2 J.
Total work = W₁ + W₂ = 1212.4 + (-606.2) = 606.2 J.
Total work done by the force = 606.2 J.

Q10: An object of mass 10 kg has a potential energy of 2940 J at a certain height. If the acceleration due to gravity is 9.8 m/s², calculate the height.
Sol: Use P.E. = m g h.
Given: P.E. = 2940 J, m = 10 kg, g = 9.8 m/s².
2940 = 10 × 9.8 × h = 98 h.
h = 2940 / 98 = 30 m.
Height = 30 m.