A 100 kg block is started with a speed of 2.0 m
s–1 on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20.
(a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt.
(b) Consider the situation com a frame of reference moving at 2.0 m s–1 along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 ms–1. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt.
(c) Find the work done in this frame by the external force holding the belt.
Given
Mass of block m=100kg
Initial velocity u=2 m/s
Coefficient of kinetic friction μ = 0.20
a) Since the block comes to stop final velocity will be zero.
So, final velocity v=0 m/s
When the block is moving over belt there is kinetic friction between the lower surface of the block and upper surface of the belt. And we know that heat is produced due to friction between two surfaces. Now because of this heat, the internal energy of block will change.
So,
Kinetic energy lost in heat due to friction = change in the internal energy
Kinetic energy lost = initial kinetic energy- final kinetic energy
change in internal energy is 200J.
b) Given
The velocity of the frame of reference uo= 2m/s
So, in this frame of reference initial and final velocity of the block will change.
New initial velocity u’=u-uo =2-2 = 0m/s
New final velocity v’ = v-uo =0-2 =-2m/s
So, increase in kinetic energy = (final – initial) kinetic energy
Increase in kinetic energy in com frame of reference is 200J.
c) Total work done in com frame of reference will be work done due to friction plus work done to give final velocity.
We know that force of friction f=
N
Where 
= coefficient of friction =0.02
N=normal reaction =mg
f = 0.02×100×10 = 200N
from newton’s second law of motion
force= mass ×acceleration
so,
200=100×acceleration
Using the third equation of motion,
v’2-u’2=2as
where s=displacement as seen in com frame of reference
(-2)2-0=2×2×s
s=1m
work done due to friction Wf =force ×displacement
Wf=200×1=200J
Now to calculate work done to give final velocity, we will work-energy theorem.
According to work-energy theorem,
Work done = change in kinetic energy
So,
Total work done W=W’ + Wf =200+200=400J.
∴ work done in com frame of reference= 400J.