Answer :

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.


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