Class 9 Work and Energy Numericals
Work and energy solved numericals
Numerical 1: Work Done by a Constant Force
Q1. A force of 12 N acts on an object and moves it through a distance of 4 m in the direction of the force. Find the work done.
Solution:
Given:
Force, F = 12 N
Displacement, s = 4 m
Formula:
Work done, W = F × s
Calculation:
W = 12 × 4
W = 48 J
Answer: Work done = 48 joules
Numerical 2: Zero Work Done
Q2. A person pushes a wall with a force of 200 N for 5 minutes, but the wall does not move. Find the work done.
Solution:
Displacement, s = 0 (wall does not move)
Formula:
W = F × s
W = 200 × 0 = 0 J
Answer: Work done = 0 joule
Because displacement is zero
Numerical 3: Work Done Against Gravity
Q3. A boy lifts a bag of mass 10 kg to a height of 2 m. Calculate the work done.
(Take g = 10 m/s²)
Solution:
Given:
m = 10 kg
h = 2 m
g = 10 m/s²
Formula:
W = mgh
Calculation:
W = 10 × 10 × 2
W = 200 J
Answer: Work done = 200 joules
Numerical 4: Kinetic Energy
Q4. Find the kinetic energy of a car of mass 1000 kg moving with a speed of 20 m/s.
Solution:
Given:
m = 1000 kg
v = 20 m/s
Formula:
KE = ½ mv²
Calculation:
KE = ½ × 1000 × (20)²
KE = 500 × 400
KE = 2,00,000 J
Answer: Kinetic energy = 2 × 10⁵ J
Numerical 5: Change in Kinetic Energy
Q5. A ball of mass 5 kg is initially at rest. It starts moving with a speed of 6 m/s. Find the work done on the ball.
Solution:
Initial velocity, u = 0
Final velocity, v = 6 m/s
Mass, m = 5 kg
Formula:
Work done = Change in kinetic energy
KE = ½ mv²
KE = ½ × 5 × 36
KE = 90 J
Answer: Work done = 90 joules
Numerical 6: Potential Energy
Q6. Calculate the potential energy of an object of mass 8 kg placed at a height of 5 m.
(g = 10 m/s²)
Solution:
Given:
m = 8 kg
h = 5 m
g = 10 m/s²
Formula:
PE = mgh
Calculation:
PE = 8 × 10 × 5
PE = 400 J
Answer: Potential energy = 400 joules
Numerical 7: Law of Conservation of Energy
Q7. A stone of mass 2 kg is dropped from a height of 10 m.
Find its kinetic energy just before hitting the ground.
(g = 10 m/s²)
Solution:
Initial potential energy = mgh
PE = 2 × 10 × 10 = 200 J
According to law of conservation of energy,
Final KE = Initial PE
Answer: Kinetic energy = 200 joules
Numerical 8: Power
Q8. A man lifts a load of 500 N to a height of 4 m in 10 seconds. Find his power.
Solution:
Work done = Force × Height
W = 500 × 4 = 2000 J
Time, t = 10 s
Formula:
Power = Work / Time
P = 2000 / 10
P = 200 W
Answer: Power = 200 watts
Numerical 9: Electrical Energy Consumption
Q9. An electric bulb of 100 W glows for 5 hours. Find the energy consumed in joules.
Solution:
Power = 100 W
Time = 5 hours = 5 × 3600 = 18000 s
Formula:
Energy = Power × Time
Energy = 100 × 18000
Energy = 18,00,000 J
Answer: Energy consumed = 1.8 × 10⁶ joules
Numerical 10: Velocity from Kinetic Energy
Q10. The kinetic energy of an object is 450 J and its mass is 10 kg. Find its velocity.
Solution:
KE = ½ mv²
450 = ½ × 10 × v²
450 = 5v²
v² = 90
v = √90 ≈ 9.5 m/s
Answer: Velocity ≈ 9.5 m/s
Exam Central
Numerical 11: Combined Concept (PE → KE)
Q11. A stone of mass 2 kg is thrown vertically upward with a speed of 10 m/s.
Find the maximum height reached by the stone.
(Take g = 10 m/s²)
Solution:
Given:
m = 2 kg
u = 10 m/s
g = 10 m/s²
v = 0 (at maximum height)
Using energy conservation:
Initial KE = Final PE
½mu² = mgh
½ × 2 × (10)² = 2 × 10 × h
100 = 20h
h = 5 m
Answer: Maximum height = 5 metres
Numerical 12: Work Done by Retarding Force
Q12. A car of mass 1200 kg moving at 20 m/s is brought to rest by applying brakes.
Find the work done by the braking force.
Solution:
Given:
m = 1200 kg
u = 20 m/s
v = 0
Work done = Change in kinetic energy
W = ½m(v² − u²)
W = ½ × 1200 × (0 − 400)
W = −240000 J
Answer: Work done = –2.4 × 10⁵ J
Negative sign shows work done against motion
Numerical 13: Power While Climbing Stairs (Very Important)
Q13. A student of mass 50 kg climbs a staircase of 30 steps, each step being 20 cm high, in 12 seconds.
Find the power developed by the student.
(Take g = 10 m/s²)
Solution:
Mass, m = 50 kg
Height of one step = 20 cm = 0.2 m
Total height = 30 × 0.2 = 6 m
Weight = mg = 50 × 10 = 500 N
Work done = mgh
W = 50 × 10 × 6 = 3000 J
Time, t = 12 s
Power = Work / Time
P = 3000 / 12
P = 250 W
Answer: Power developed = 250 watts
Numerical 14: Energy at Halfway Point
Q14. An object of mass 4 kg is dropped from a height of 20 m.
Find its kinetic energy when it is at a height of 10 m above the ground.
(Take g = 10 m/s²)
Solution:
Initial potential energy = mgh
= 4 × 10 × 20
= 800 J
Potential energy at 10 m height:
= 4 × 10 × 10
= 400 J
Using conservation of energy:
KE = Total energy − Remaining PE
KE = 800 − 400
KE = 400 J
Answer: Kinetic energy = 400 joules
Numerical 15: Multiple Appliances Energy Consumption
Q15. Three electrical appliances of power 500 W, 1000 W, and 1500 W work for 2 hours each.
Find the total energy consumed in kilowatt-hour (kWh) and in joules.
Solution:
Total power = 500 + 1000 + 1500
Total power = 3000 W = 3 kW
Time = 2 hours
Energy in kWh = Power × Time
Energy = 3 × 2
Energy = 6 kWh
Conversion:
1 kWh = 3.6 × 10⁶ J
Energy in joules = 6 × 3.6 × 10⁶
= 21.6 × 10⁶ J
Answer:
Energy consumed = 6 kWh
= 2.16 × 10⁷ joules
