A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-
Solution
$P = F.v = mav$<br><br>$P = {{mvdv} \over {dt}}$<br><br>$\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv}$<br><br>$Pt = {{m{v^2}} \over 2}$<br><br>$v = \sqrt {{{2Pt} \over m}}$<br><br>${{dx} \over {dt}} = \sqrt {{{2Pt} \over m}}$<br><br>$\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt$<br><br>$x \propto {t^{3/2}}$
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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