Here we are with 24 important numerical problems for class 9 chapter 10 Work and Energy. Below are the important formulas that would be used for solving the numericals below.
Formulas
- Work (force along displacement): W = F × s
- 1 joule: 1 J = 1 N × 1 m
- Kinetic energy: Ek = (1/2) × m × v²
- Potential energy at height: Ep = m × g × h
- Work–energy theorem: Work done = change in kinetic energy = Ekf − Eki
- Power: P = W ÷ t
- 1 watt: 1 W = 1 J/s
- 1 kW = 1000 W
- 1 unit of electricity = 1 kWh = 3.6 × 10⁶ J
- Speed conversion: km/h to m/s → multiply by 5/18
- Conservation of mechanical energy (free fall, ignoring air resistance): Ep + Ek = constant
Class 9 Science (NCERT) – Chapter 10: Work and Energy
Basic
Q1) A force of 7 N acts on an object and produces a displacement of 8 m in the direction of the force. Find the work done.
Solution:
Given: F = 7 N, s = 8 m
Use W = F × s
W = 7 × 8 = 56 J
Answer: 56 J
Q2) A pair of bullocks exerts a force of 140 N on a plough. The field is 15 m long. Find the work done in ploughing the field.
Solution:
Given: F = 140 N, s = 15 m
Use W = F × s
W = 140 × 15 = 2100 J
Answer: 2100 J
Q3) An object of mass 15 kg is moving with a speed of 4 m/s. Find its kinetic energy.
Solution:
Given: m = 15 kg, v = 4 m/s
Use Ek = (1/2) × m × v²
Ek = (1/2) × 15 × 4²
Ek = (1/2) × 15 × 16
Ek = 120 J
Answer: 120 J
Q4) Find the potential energy of a 10 kg object kept at a height of 6 m. Take g = 9.8 m/s².
Solution:
Given: m = 10 kg, h = 6 m, g = 9.8 m/s²
Use Ep = m × g × h
Ep = 10 × 9.8 × 6
Ep = 588 J
Answer: 588 J
Q5) A lamp uses 1000 J of energy in 10 s. Find its power.
Solution:
Given: W = 1000 J, t = 10 s
Use P = W ÷ t
P = 1000 ÷ 10 = 100 W
Answer: 100 W
Q6) A force of 5 N acts on an object, but the object does not move at all. Find the work done.
Solution:
Given: displacement s = 0 m
Use W = F × s
W = 5 × 0 = 0 J
Answer: 0 J
Standard
Q7) A porter lifts a luggage of mass 15 kg to a height of 1.5 m. Take g = 10 m/s². Find the work done by him.
Solution:
Given: m = 15 kg, h = 1.5 m, g = 10 m/s²
Force needed = weight = m × g = 15 × 10 = 150 N
Work done W = force × displacement
W = 150 × 1.5 = 225 J
Answer: 225 J
Q8) A car of mass 1500 kg increases its speed from 30 km/h to 60 km/h. Find the work done to increase its speed.
Solution:
Given: m = 1500 kg, u = 30 km/h, v = 60 km/h
Convert to m/s:
u = 30 × (5/18) = 25/3 m/s
v = 60 × (5/18) = 50/3 m/s
Initial kinetic energy:
Eki = (1/2) × m × u²
Eki = (1/2) × 1500 × (25/3)²
Eki = 750 × (625/9)
Eki = 468750/9 J
Eki = 156250/3 J
Final kinetic energy:
Ekf = (1/2) × 1500 × (50/3)²
Ekf = 750 × (2500/9)
Ekf = 1875000/9 J
Ekf = 625000/3 J
Work done = Ekf − Eki
Work done = (625000/3) − (156250/3)
Work done = 468750/3
Work done = 156250 J
Answer: 156250 J
Q9) A 12 kg object has potential energy 480 J. Take g = 10 m/s². Find the height.
Solution:
Given: m = 12 kg, Ep = 480 J, g = 10 m/s²
Use Ep = m × g × h
480 = 12 × 10 × h
480 = 120 × h
h = 480 ÷ 120 = 4 m
Answer: 4 m
Q10) A force changes the speed of a 20 kg object from 5 m/s to 2 m/s. Find the work done by the force.
Solution:
Given: m = 20 kg, initial speed u = 5 m/s, final speed v = 2 m/s
Work done = change in kinetic energy = Ekf − Eki
Eki = (1/2) × m × u²
Eki = (1/2) × 20 × 5²
Eki = 10 × 25 = 250 J
Ekf = (1/2) × 20 × 2²
Ekf = 10 × 4 = 40 J
Work done = 40 − 250 = −210 J
Answer: −210 J
Q11) A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. Height of each step is 15 cm. Take g = 10 m/s². Find his power.
Solution:
Given: m = 50 kg, steps = 45, height per step = 15 cm = 0.15 m, t = 9 s, g = 10 m/s²
Total height h = 45 × 0.15 = 6.75 m
Work done W = m × g × h
W = 50 × 10 × 6.75 = 3375 J
Power P = W ÷ t
P = 3375 ÷ 9 = 375 W
Answer: 375 W
Q12) A house consumed 250 units of electricity in a month. Find the energy in joules.
Solution:
Given: 250 units = 250 kWh
1 kWh = 3.6 × 10⁶ J
Energy = 250 × 3.6 × 10⁶ J
Energy = 900 × 10⁶ J
Energy = 9 × 10⁸ J
Answer: 9 × 10⁸ J
Advance
Q13) A 40 kg object is raised to a height of 5 m. Take g = 10 m/s². Find its potential energy.
Solution:
Given: m = 40 kg, h = 5 m, g = 10 m/s²
Ep = m × g × h
Ep = 40 × 10 × 5 = 2000 J
Answer: 2000 J
Q14) The same 40 kg object of Q13 is allowed to fall. Find its kinetic energy when it is half-way down.
Solution:
From Q13: Total mechanical energy at top = Ep(top) = 2000 J
Half-way down means height = 2.5 m
Potential energy at half-way:
Ep(half) = m × g × h
Ep(half) = 40 × 10 × 2.5 = 1000 J
By conservation of energy:
Ep + Ek = constant
Ek(half) = 2000 − 1000 = 1000 J
Answer: 1000 J
Q15) Calculate the work required to stop a car of mass 1500 kg moving at 60 km/h.
Solution:
Given: m = 1500 kg, v = 60 km/h
Convert speed:
v = 60 × (5/18) = 50/3 m/s
To stop the car, final kinetic energy = 0
Work required = change in kinetic energy = 0 − Ek(initial)
Ek(initial) = (1/2) × m × v²
Ek = (1/2) × 1500 × (50/3)²
Ek = 750 × (2500/9)
Ek = 1875000/9 J
Ek = 625000/3 J
Work required = −625000/3 J
Answer: −625000/3 J
Q16) An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Solution:
Given: P = 1500 W, time = 10 hours
Convert time to seconds:
10 hours = 10 × 3600 = 36000 s
Energy used W = P × t
W = 1500 × 36000
W = 54000000 J
W = 5.4 × 10⁷ J
Answer: 5.4 × 10⁷ J
Q17) Find the energy used in 10 hours by four devices of power 500 W each.
Solution:
Given: power of each device = 500 W, number of devices = 4, time = 10 hours
Total power = 4 × 500 = 2000 W
Time in seconds = 10 × 3600 = 36000 s
Energy W = P × t
W = 2000 × 36000
W = 72000000 J
W = 7.2 × 10⁷ J
Answer: 7.2 × 10⁷ J
Q18) Two girls each have weight 400 N and climb a height of 8 m. Girl A takes 20 s and Girl B takes 50 s. Find the power of each.
Solution:
Given: weight = 400 N, height h = 8 m
Work done by each = force × displacement
W = 400 × 8 = 3200 J
Power of A:
PA = W ÷ t
PA = 3200 ÷ 20 = 160 W
Power of B:
PB = 3200 ÷ 50 = 64 W
Answer: Girl A = 160 W, Girl B = 64 W
HOTS
Q19) A ball is dropped from a height of 10 m. After striking the ground, its energy reduces by 40%. Take g = 10 m/s². Find the height to which it bounces back.
Solution:
Initial potential energy at 10 m = m × g × 10
After collision, energy left = 60% of initial energy
Energy left = 0.6 × (m × g × 10)
At maximum bounce height H, this energy becomes potential energy:
m × g × H = 0.6 × (m × g × 10)
Cancel m × g on both sides:
H = 0.6 × 10
H = 6 m
Answer: 6 m
Q20) A rocket is moving up with speed v. If its speed becomes 3v, find the ratio of the two kinetic energies.
Solution:
Initial kinetic energy:
E1 = (1/2) × m × v²
Final kinetic energy:
E2 = (1/2) × m × (3v)²
E2 = (1/2) × m × 9v²
Ratio E2 : E1 = 9v² : v² = 9 : 1
Answer: 9 : 1
Q21) A motor pump has power 2 kW. How much water can it raise per minute to a height of 10 m? Take g = 10 m/s².
Solution:
Given: P = 2 kW = 2000 W, h = 10 m, g = 10 m/s², time = 1 minute = 60 s
Power = work ÷ time
Work in 60 s = P × t
Work = 2000 × 60 = 120000 J
Work used to lift water = m × g × h
120000 = m × 10 × 10
120000 = 100m
m = 1200 kg
Answer: 1200 kg of water per minute
Q22) A 1000 kg car moves on a level road at 36 km/h against a friction force of 100 N. Find the power needed.
Solution:
Given: friction force F = 100 N, speed = 36 km/h
Convert speed:
v = 36 × (5/18) = 10 m/s
Power needed to overcome friction:
P = F × v
P = 100 × 10 = 1000 W
Answer: 1000 W
Q23) A girl of mass 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial speed 4 m/s and comes to rest after moving 16 m. Find the total work done on the trolley system.
Solution:
Total mass (girl + trolley) = 35 + 5 = 40 kg
Initial speed u = 4 m/s, final speed v = 0 m/s
Work done = change in kinetic energy = Ekf − Eki
Eki = (1/2) × 40 × 4²
Eki = 20 × 16 = 320 J
Ekf = 0
Work done = 0 − 320 = −320 J
Answer: −320 J
Q24) Jog Falls are 20 m high. 2000 tonnes of water fall in 1 minute. Take g = 10 m/s². Find the equivalent power if all this energy is used.
Solution:
Given: height h = 20 m, mass of water = 2000 tonnes, time = 1 minute
Convert mass:
1 tonne = 1000 kg
Mass m = 2000 × 1000 = 2000000 kg
Time t = 60 s
Energy released (work done by gravity):
W = m × g × h
W = 2000000 × 10 × 20
W = 400000000 J
W = 4 × 10⁸ J
Power:
P = W ÷ t
P = (4 × 10⁸) ÷ 60
P = (2 × 10⁸) ÷ 30
P = 6666666.67 W
Answer: 6666666.67 W
