Numerical Questions – Chapter: Energy
1. A motorbike of mass 140 kg is moving at a speed of 30 km/h. Find its kinetic energy.
Answer:
First convert speed to m/s:
30 km/h = 30 × (1000/3600) = 8.33 m/s
K.E. = (1/2)mv² = (1/2) × 140 × (8.33)² = 7000 J
✅ Answer: 7,000 J
First convert speed to m/s:
30 km/h = 30 × (1000/3600) = 8.33 m/s
K.E. = (1/2)mv² = (1/2) × 140 × (8.33)² = 7000 J
✅ Answer: 7,000 J
2. A boy of mass 40 kg is running at 1 m/s inside his classroom, which is on the second floor of his school. If each floor is 12 m above the one below it, find his:
a) Kinetic Energy
b) Potential Energy
a) Kinetic Energy
b) Potential Energy
Answer:
a) K.E. = (1/2)mv² = (1/2) × 40 × (1)² = 20 J
b) P.E. = mgh = 40 × 10 × 24 = 9,600 J
✅ Answers: K.E. = 20 J; P.E. = 9,600 J
a) K.E. = (1/2)mv² = (1/2) × 40 × (1)² = 20 J
b) P.E. = mgh = 40 × 10 × 24 = 9,600 J
✅ Answers: K.E. = 20 J; P.E. = 9,600 J
3. Consider an object of mass m moving with a kinetic energy K. If its velocity is doubled, how much should the mass be decreased so that the kinetic energy does not change?
Answer:
K.E. ∝ mv²
If velocity becomes 2v, to keep K.E. constant:
m × (2v)² = original mv² → m × 4v² = constant → New mass = m/4
✅ Answer: Mass should be reduced to one-fourth (4 times less)
K.E. ∝ mv²
If velocity becomes 2v, to keep K.E. constant:
m × (2v)² = original mv² → m × 4v² = constant → New mass = m/4
✅ Answer: Mass should be reduced to one-fourth (4 times less)
