NEET-XII-Physics
26: Laws of Thermodynamics
- #3A 100 kg lock 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.Ans : Here,
m = 100 kg
u = 2.0 m/s
v = 0
μk = 0.2
a) Internal energy of the belt-block system will decrease when the block will lose its KE in heat due to friction. Thus,
KE lost =
b) Velocity of the frame is given by
uf = 2.0 m/s
u’ = u - uf = 2 - 2= 0
v’ = 0 - 2 = -2 m/s
KE lost =
c) Force of friction is given by
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