A block of mass 2 kg moving on a horizontal surface with speed of 4 ms$-$1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = $-$kx where k = 12 Nm$-$1. The speed of the block as it just crosses the rough surface will be :
Solution
<p>$F = - 12x$</p>
<p>$mv{{dv} \over {dx}} = - 12x$</p>
<p>$\int_4^v {vdv = - 6\int_{0.5}^{1.5} {xdx} }$ ($m = 2$ kg)</p>
<p>${{{v^2} - 16} \over 2} = - 6\left[ {{{{{1.5}^2} - {{0.5}^2}} \over 2}} \right]$</p>
<p>${{{v^2} - 16} \over 2} = - 6$</p>
<p>$v = 2$ m/sec</p>
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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