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A block of mass 2 kg moving on a horizontal surface with speed of $\mathrm{4 \mathrm{~ms}^{-1}}$ enters a rough surface ranging from $\mathrm{x=0.5 \mathrm{~m} \text { to } x=1.5 \mathrm{~m} \text {. }}$ The retarding force in this range of rough surface is related to distance by $\mathrm{\mathrm{F}=-\mathrm{k} x \text { where } \mathrm{k}=12 \mathrm{Nm}^{-1}}$ . The speed of the block as it just crosses the rough surface will be :
Option: 1
$\mathrm{\text { zero }}$

Option: 2
$\mathrm{1.5 \mathrm{~ms}^{-1}}$

Option: 3
$\mathrm{2.0 \mathrm{~ms}^{-1}}$

Option: 4
$\mathrm{2.5 \mathrm{~ms}^{-1}}$

Answers (1)

best_answer

$\mathrm{F=-12 x}$

$\mathrm{mv \frac{d v}{d x}=-12x}$

$\mathrm{v d v=-\frac{12}{m} x d x }$

$\mathrm{\int_{v=4}^{v}vdv=-6\int_{x=0.5}^{x=1.5}xdx }$

$\mathrm{ \left [\frac{v^{2}}{2} \right ] ^{v}_{4}}=\mathrm{-6\left [ \frac{x^{2}}{2} \right ]^{1.5}_{0.5}}$

$\mathrm{\frac{v^{2}-16}{2}=-\frac{6}{2}\left[1.5^{2}-0.5^{2}\right]}$

$\mathrm{\frac{v^{2}}{2}-8=-3\times 2}$

$\mathrm{\frac{v^{2}}{2}=8-6=2}$

$\mathrm{\frac{v^{2}}{2}=8-6=2}$

$\mathrm{v^{2}=4}$

$\mathrm{v=2 m / s}$

Hence 3 is the correct answer.

Posted by

vinayak

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