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A block of mass $15 \mathrm{~kg}$ 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 $0.5$ $\mathrm{ms}^{-2}$ for $20 \mathrm{~s}$ and then moves with a 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.
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Given, $m=15 \mathrm{~kg}, \mu=0.18, a=0.5 \mathrm{~ms}^{-2}, t=20 \mathrm{~s}$, Force acting on the block due to the motion of the trolley, $F^{\prime}=m a=15 \times 0.5=7.5 \mathrm{~N}$, which acts in the forward direction.
Force due to the liming friction on the block, $F=\mu R=\mu m g=0.18 \times 15 \times 9.8=26.46 \mathrm{~N}$, which opposes the motion of the block
$\because F^{\prime} < F$, the block will not move
$\therefore$ (a) for stationary observer, the block would be at rest with respect to the trolley.
(b) When the observer moves with an acceleration with the trolley, he would be in a non-inertial frame, where law of inertia would not be valid. The box will be at rest relative to the observer.
Force due to the liming friction on the block, $F=\mu R=\mu m g=0.18 \times 15 \times 9.8=26.46 \mathrm{~N}$, which opposes the motion of the block
$\because F^{\prime} < F$, the block will not move
$\therefore$ (a) for stationary observer, the block would be at rest with respect to the trolley.
(b) When the observer moves with an acceleration with the trolley, he would be in a non-inertial frame, where law of inertia would not be valid. The box will be at rest relative to the observer.
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