JEE Class main Answered
Sir pls solve the following.
Asked by rsudipto | 22 Dec, 2018, 08:04: PM
Expert Answer
Left side figure shows the free body diagram of the block that shows the different forces acting on 4 kg block at t=0.
Both applied force 2N and friction force μR = 0.1×4g = 4N are acting in the opposite direction of movement.
Hence initial motion of the block is retarded motion. Retardation = 6/4 = 1.5 m/s² .
The block will come to rest after time duration = 3/1.5 = 2 seconds
The distance moved in 2 seconds S = ut-(1/2)at² = 3×2 - (1/2)×1.5×2×2 = 3 m
after t=2 s, the block comes to rest, but the applied force which is acting in the opposite dirtection will try to move the block towards left.
Hence the friction force starts acting towards right. This is shown in the right side of figure.
But friction force 4N is greater than applied force 2N, hence block willnot move.
Hence after 5 s, distance moved is 3 m towards right, i.e., 3
Answered by Thiyagarajan K | 23 Dec, 2018, 07:51: AM
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