Q. 34.5( 2 Votes )
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.
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.
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
Using the third equation of motion,
where s=displacement as seen in com frame of reference
work done due to friction Wf =force ×displacement
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
Total work done W=W’ + Wf =200+200=400J.
∴ work done in com frame of reference= 400J.
Rate this question :
50 cal of hHC Verma - Concepts of Physics Part 2
Consider two procHC Verma - Concepts of Physics Part 2
The final volumeHC Verma - Concepts of Physics Part 2