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Question

Two blocks A and B of mass 2kg and 4kg respectively are placed one over the other as shown in the figure. A time varying horizontal force F=2t is applied on the upper block as shown in figure. Here t is in second and F is in newton. Draw a graph showing accelerations of A and B. The coefficient of friction between A and B is μ=12 and the horizontal surface over which B is placed is smooth. (g=10m/s2)
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Solution

Limiting friction between A and B,

fl = μmag = 12(2)(10) = 10N

Block B moves due to friction only, therefore maximum acceleration of B can be

amax = flmb = 104 = 2.5 m/s2

Thus both the blocks move together with same acceleration till the common acceleration becomes 2.5 m/s2, after that acceleration of B will become constant while that of A will go on increasing. To find the time when the acceleration of both the blocks becomes 2.5m/s2 (or when slipping will start between A and B) we will write,

2.5 = Fma+mb = 2t6

t = 7.5 sec

Hence for t <= 7.5 S,

aa = ab = t3

Thus aA versus t or aB versus t graph is a straight line passing through origin of slope 1/3. Hence the graph is shown above.



751239_244825_ans_a2622abbd4264fc8a4adcf1379baecf0.PNG

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