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Byju's Answer

Standard XII

Physics

Maximum Push Using Friction

A bar of mass...

Question

A bar of mass $m_{1}$ is placed on a plank of mass $m_{2}$ , which rests on a smooth horizontal plane. The coefficient of friction between the surfaces of bar and plank is k. The plank is subjected to a horizontal force F depending on time t as F = at, where a is a constant. The moment of time $t_{0}$ at which the plank starts sliding relative to bar is

\frac{a k g}{m_{1} + m_{2}}

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\frac{(m_{1} + m_{2}) k g}{a}

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\frac{(m_{1} + m_{2}) g}{k a}

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\frac{k a}{(m_{1} + m_{2}) g}

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Solution

The correct option is B $\frac{(m_{1} + m_{2}) k g}{a}$
$a_{m a x} = μ g$
so the force at an instant when the slipping starts is
$F = (m_{1} + m_{2}) a_{m a x} = (m_{1} + m_{2}) μ g$
$a t_{0} = (m_{1} + m_{2}) μ g ⟹ t_{0} = \frac{(m_{1} + m_{2}) k g}{a}$

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The correct option is B (m1+m2)kgaamax=μgso the force at an instant when the slipping starts is F=(m1+m2)amax=(m1+m2)μgat0=(m1+m2)μg⟹t0=(m1+m2)kga

The correct option is B $\frac{(m_{1} + m_{2}) k g}{a}$
$a_{m a x} = μ g$
so the force at an instant when the slipping starts is
$F = (m_{1} + m_{2}) a_{m a x} = (m_{1} + m_{2}) μ g$
$a t_{0} = (m_{1} + m_{2}) μ g ⟹ t_{0} = \frac{(m_{1} + m_{2}) k g}{a}$