A block of mass m slides from rest at a heigh | vedclass

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Question

(A) The block starts from rest at height $H$,so its initial mechanical energy is $E_i = mgH$.As the block moves,it encounters a rough horizontal surfa

Vedclass · Accepted Answer

$h=H-2 \mu(d+x)$

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(A) The block starts from rest at height $H$,so its initial mechanical energy is $E_i = mgH$.As the block moves,it encounters a rough horizontal surface of length $d$ and then compresses the spring by distance $x$. The work done against friction during the forward journey is $W_{f1} = \mu mgd$.During the compression of the spring,the block travels an additional distance $x$ on the rough surface,so the work done against friction during this part is $W_{f2} = \mu mgx$.The total work done against friction during the forward journey is $W_{f,total} = \mu mg(d+x)$.When the block returns,it travels the same distance $x$ (spring expansion) and $d$ on the rough surface,so the work done against friction during the return journey is also $W_{f,return} = \mu mg(d+x)$.By the work-energy theorem,the initial potential energy equals the total work done against friction plus the final potential energy $mgh$ at height $h$:$mgH = W_{f,total} + W_{f,return} + mgh$$mgH = \mu mg(d+x) + \mu mg(d+x) + mgh$$mgH = 2\mu mg(d+x) + mgh$Dividing by $mg$,we get:$H = 2\mu(d+x) + h$$h = H - 2\mu(d+x)$

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