## Work, Energy and Power

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

*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 10

^{6}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/s**

^{2})Answer:

Given

acceleration (a) = g = 10 m/s

^{2}

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/s**

^{2})Answer: Given,

a = g = - 10 m/s

^{2}

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 mv

⇒ KE (final) = 1.44 KE (initial)^{2}= ½ m(1.2u)^{2 }= ½ m × 1.44u^{2}**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 mv

^{2}= F × s

⇒ F = ½ mv

^{2 }× (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/s**

^{2})
Answer: Given,

mass m = 60 kg

height h = 8848 m

g = 9.8 m/s

^{2}
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/s**

^{2})
Answer: Given mass (m) = 20 kg

displacement (h) = 10m

g = 10 m/s

^{2}
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?**

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