A block is placed on a rough horizontal plane | vedclass

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Question

Block starts sliding over the surface when, $\mathrm{F}=\max .$ force of friction

Or      $\mathrm{Kt}=\mu \mathrm{mg}$

Or

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Vedclass · Answer

Block starts sliding over the surface when, $\mathrm{F}=\max .$ force of friction

Or      $\mathrm{Kt}=\mu \mathrm{mg}$

Or      $\mathrm{t}=\mu \mathrm{mg} / \mathrm{K}$

For $\mathrm{t}>\mu \mathrm{mg} / \mathrm{K}:$

net acceleration of block is,

$\mathrm{a}=\frac{\mathrm{F}-\mu \mathrm{mg}}{\mathrm{m}}=\frac{\mathrm{Kt}}{\mathrm{m}}-\mu \mathrm{g}$

which is a straight line with positive slope and negative intercept.

So, for $t \leq \mu m g / K,$ acceleration of the particle is zero

and for $\mathrm{t}>\mu \mathrm{mg} / \mathrm{K},$ acceleration is $\mathrm{a}=| \mathrm{Kt} / \mathrm{m}-$

$\mu g$

A block is placed on a rough horizontal plane. A time dependent horizontal force $F = Kt$ acts on the block. Here $K$ is a positive constant. Acceleration-time graph of the block is

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