A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
Soln:
Kinetic energy equation: Ek=1/2 mv2
Momentum equation: p=mv
Kinetic energy equation in terms of momentum: Ek = p2/2m
Kinetic energy of two objects will be the same if their mass and momentum are the same. Among light and heavy bodies, Heavy body has highest momentum. When the momentum is the equal light body has more kinetic energy.
22. An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.
Soln:
m(A)= m(B) = 1000 kg.
v = 36 km/h =10 m/s
Frictional force = 100 N
car A moves with a uniform speed, which means engine of car applies a force equal to the frictional force
Power = Force × distance / time = F . V
= 100 N × 10 m/s = 1000 W
After collision mA uA + mB uB = mA vA + mB vB.
1000 × 10 + 1000 × 0
= 1000 × 0 + 1000 × vB vB
= 10 m s–1
23. A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?
Soln:
Initial velocity of the trolley,u = 4 m/s
Final velocity of the trolley v = 0
Mass of the trolley m = 5 kg
Distance covered by the trolley before coming to rest,s = 16 m
From Equation 2 as = v2-u2,
a = v2-u22 S
= 0-(4)22×16
= 0.5 m/s2
Force (frictional) acting on the trolley = ma
= 40 (- 0.5)
= – 20 N
Work done on the trolley = Fs = (20 N) (16 m)
= 320 J
(b) Since the girl does not move w.r.t. the trolley (as she is sitting on it), work done by the girl = 0.
24. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.
(a) How much work is done by the men in lifting the box?
(b) How much work do they do in just holding it?
(c) Why do they get tired while holding it? (g = 10 m s–2)
Soln:
(a)
F= 250kg x g = 2500 N
S= 1m
W=F.s= 250Nm = 2500 J
(b)
While holding box there is no displacement hence work done is zero
c) While holding the box they apply a muscular force which is equal and opposite to the gravitational force hence they feel tired while holding box.
25. What is power? How do you differentiate kilowatt from kilowatt-hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tonnes of waterfalls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 m s–2)
Soln:
Power is the rate of transfer of energy or the rate of doing work. Watt is the unit of power and kilowatt is 1000 watts.
h = 20 m, and mass = 2000 × 103 kg
= 2 × 106 kg
Power = m g h / t
= 2 x 106 x 10 x 20 / 60
w = 4/6 x 107
w = 2/3 x 107w
26. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at a constant speed of 1 m s –1 vertically? (g = 10 m s–2)
Soln:
Power = work/time
work = force x displacement
force = mass x acceleration
acceleration = velocity/time
Therefore
Power = velocity × mass × displacement / time × time
Here
Power,P=100W
velocity,v=1m/s
since time,t=1s
displacement, s=1m
acceleration,a=10m/s
From equation P=m×a×s / t
100=m×10
m=10
27. Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?
Soln:
Power of an object which does work at 1 Joule/sec is called watt.
1 watt =
1 kilowatt = 1000watts = 1000J/sec
mass of car = 150kg
power for each kg = 500x 150=7500w
speed = 20m/s
power = Force x v
∴force=power/v = 7500/20
=350N
28. Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 m s–2) (i) a butterfly of mass 1.0 g that flies upward at a rate of 0.5 m s–1. (ii) a 250 g squirrel climbing up on a tree at a rate of 0.5 m s–1.
Soln:
- Butterfly force= mg=1/100power=f×s÷t. {s÷t=velocity}power=1÷100×.5power=1÷200W
- squirrel force=mg=25power=25×.5power=12.5W
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