A body of mass $2 \mathrm{~kg}$ is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in $4 \mathrm{~s}$ is $\frac{1}{3} \alpha^{2} \sqrt{P} m$. The value of $\alpha$ will be ______.
Answer (integer)
4
Solution
<p>$P = Fv$</p>
<p>$m{{vdv} \over {dt}} = P$</p>
<p>$m\int_0^v {vdv = \int_0^t {Pdt} }$</p>
<p>${{m{v^2}} \over 2} = Pt$</p>
<p>$v = \sqrt {{{2P} \over m}} {t^{1/2}}$</p>
<p>$\int_0^s {dx = \sqrt {{{2P} \over m}} \int_0^t {{t^{1/2}}dt} }$</p>
<p>$s = {2 \over 3}\sqrt {{{2P} \over m}} {t^{3/2}}$</p>
<p>or $s = {2 \over 3}\sqrt {{{2P} \over 2}} \times {4^{3/2}}$</p>
<p>$= {{16} \over 3}\sqrt P ~m$</p>
<p>So, $\alpha = 4$</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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