A block slides down a rough inclined plane of length 250 m at a constant speed. The angle of inclination of the plane is 37^ with the horizontal. If 50 % of the heat generated due to friction is absorbed by the block, what will be the rise in the temperature of the block? [Take g = 10 m/s ^2 , specific heat capacity of the block s = 500 J/(kg ^ C) and (37^ ) = 0.6 ]

Step-by-step solution

Since the block moves at a constant speed, its kinetic energy does not change. The work done against friction is equal to the loss in gravitational potential energy. Loss in potential energy = mgh = mgL (37^ ) = m 10 250 0.6 = 1500m J Heat generated by friction = 1500m J Heat absorbed by the block, Q = 50 % of 1500m = 750m J Using the calorimetry principle Q = ms T : 750m = m 500 T T = 750 500 = 1.5^ C Answer: 1.5^ C