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Sr please solve the following
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Asked by misra.dinesh2706 | 30 Dec, 2018, 19:08: PM
answered-by-expert Expert Answer
Figure shows the forces acting on the sliding block.
Tension T in the string = mg sin37 - f  ..............(1)
where m is mass of block, g is acceleration due to gravity and f friction force.
 
Hence,  T = 0.2×9.8×0.6 - 0.5 = 0.676 N
 
Torque τ acting on the pulley = T×r .............(2)
where r is radius of pulley.
 
hence,  τ = 0.676×0.7 = 0.473 N-m
 
angular acceleration α of rotation of pulley is given by, τ = I×α ..................(3)
where I is moment of inertia.
 
hence angular acceleration α = τ / I  = 0.473 / 0.8 = 0.59 rad/s2
linear acceleration a of the block = α×r = 0.59×0.7 = 0.41 m/s2
 
speed v at the distance 100 cm is obtained from :  v = [ 2×a×s ]1/2  = [ 2×0.41×0.1 ]1/2 = 0.29 m/s
Answered by Thiyagarajan K | 30 Dec, 2018, 22:51: PM

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