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Laws of Motion

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Question : 35 of 40
Marks: +1, -0
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s−2^{-2} for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Solution:  
Here, m=15 kgm = 15\,\text{kg}; a=0.5 m s−2a = 0.5\,\text{m}\,\text{s}^{-2}, t=20 st = 20\,\text{s} and m=0.18m = 0.18
Force on the block due to the motion of the trolley,
F=ma=15×0.5=7.5 NF = ma = 15 \times 0.5 = 7.5\,\text{N}
Force of limiting friction on the block,
f′=mR=μmgf' = mR = \mu m g =0.18×15×10=27 N= 0.18 \times 15 \times 10 = 27\,\text{N}
(a) To a stationary observer on the ground, force F on the block acts so as to cause the motion and the force f opposes the motion of the block. Since f>Ff > F, the block will continue to remain stationary. In fact, the force of limiting friction f adjusts itself to be equal to the force F.
(b) The motion of the trolley is an accelerated one. Therefore, the trolley is a non-inertial frame of reference and the observer moving with the trolley is in a non-inertial frame. The laws of mechanics are no longer valid in such a frame.
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