Solve the following :


A $250 \mathrm{~g}$ block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of $40 \mathrm{~cm} / \mathrm{s}$. If the friction coefficient between the table and the block is $0.1$, how far does the block move before coming to rest?


We know,

$\mathrm{ma}=\mu \mathrm{R}$

$a=\frac{(\mu \mathrm{mg})}{m}=\mu \mathrm{g}$

$a=0.1 \times 9.8=0.98 \mathrm{~m} / \mathrm{s}^{2}$


$v^{2}-u^{2}=2 a s$

$\mathrm{s}=\frac{\mathrm{v}^{2}-\mathrm{u}^{2}}{2 a}=\left(0-0.4^{2}\right) / 2 \times 0.98$

$\mathrm{s}=0.082 \mathrm{~m}$

Work done

$W=-\mu R \sin \theta=-0.1 \times 2.5 \times 0.082 \times 1$

$W=-0.02 \mathrm{~J}$



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