## Work, Energy and Power

### Numerical Problems based on Class 9 Physics NCERT Chapter

Q1(CBSE 2012): State the relation between commercial unit of energy and joules.

Answer: Kilowatt hour (kwh)
1 kwh - 3.6 x 106 joules

Q2: How much work is done on a body of mass 1kg whirling on a circular path of radius 5m?

Answer: Zero

Q3: A body of mass 15 kg undergoes downward displacement of 40m under the effect of gravitational force. How much work is done? (take g = 10 m/s2)

Answer:
Given
acceleration (a) = g = 10 m/s2
mass (m) = 15 kg
displacement (s) = 40 m

Work W = F × s
= m × a × s = 15 × 10 × 40 = 6000 J
Since gravitational force and displacement are in same direction, work is said to be positive and done on the body.

Q4: A body of mass 120 g is taken vertically upwards to reach the height of 5m. Calculate work done. (Take g = 10 m/s2)

Answer: Given,
a = g = - 10 m/s2
m = 120g = 0.120 kg
s = 5m
Work W = F × s
= m × g × s = 0.12 × -10 × 5 = -6 J

Since Force and displacement act in opposite directions, work is negative. Here work is considered to be done by the body against the applied force.

Q5: A momentum of the body is increased by 20%. What is the percentage increase in its kinetic energy?

Answer: Let m be the mass of the body.
initial velocity = u m/s
final velocity = v m/x
Final momentum = 1.2 times of initial momentum
⇒      mv = 1.2 × mu
⇒      v = 1.2 u
⇒   KE (final) =  ½ mv2 =  ½ m(1.2u)2   = ½ m × 1.44u2
⇒   KE (final) = 1.44 KE (initial)

Q6: A 30g bullet initially travelling at 500 m/s penetrates 12 cm into a wooden block. How much average force does it exerts?

Answer: Given,
displacement (s) = 12 cm = 0.12 m
velocity (u) = 500 m/s
mass (m) = 30g = 0.03 kg

Applying work energy theorem,
KE = Work Done
⇒  ½ mv2 =  F × s
⇒  F = ½ mv× (1/s) = ½ × 0.03 × (500)2 × (1/0.12) = 31250 N

Q7: Calculate potential energy of a person having 60 kg mass on the summit of Mt. Everest. Height of Mt. Everest is 8848 m from sea level. (g = 9.8 m/s2)

Answer: Given,
mass m = 60 kg
height h = 8848 m
g = 9.8 m/s2
Potential energy U = mgh  = (60) × (9.8)× (8848) = 52,02,624 J

Q8: A boy of mass 20 kg climbs up a ladder in 20 seconds. If the height of a ladder is 10 metres, calculate amount of power used.(Take g = 10 m/s2)

Answer:  Given   mass (m) = 20 kg
displacement (h) = 10m
g = 10 m/s2
time (t) = 20s
PE = mgh = 20 × 10 × 10 = 2000 J

Power (P) = W / t
= 2000 / 20  = 100 J/s = 100 Watt = 0.1 kW

Video below shows lowering a ramp affects on the velocity. Can you state the reason?