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Question Number 24716 by Tinkutara last updated on 25/Nov/17

$$\mathrm{A}2.2\mathrm{kg}\mathrm{block}\mathrm{starts}\mathrm{from}\mathrm{rest}\mathrm{on}\mathrm{a}$$$$\mathrm{rough}\mathrm{inclined}\mathrm{plane}\mathrm{that}\mathrm{makes}\mathrm{an}$$$$\mathrm{angle}\mathrm{of}25°\mathrm{with}\mathrm{the}\mathrm{horizontal}.\mathrm{The}$$$$\mathrm{coefficient}\mathrm{of}\mathrm{kinetic}\mathrm{friction}\mathrm{is}0.25.\mathrm{As}$$$$\mathrm{the}\mathrm{block}\mathrm{goes}2\mathrm{m}\mathrm{down}\mathrm{the}\mathrm{plane},\mathrm{the}$$$$\mathrm{mechanical}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{Earth}-\mathrm{block}$$$$\mathrm{system}\mathrm{changes}\mathrm{by}$$

Answered by ajfour last updated on 25/Nov/17

$$△U=mg△y$$$$△K=\frac{1}{2}{mv}^2$$$$△E=△U+△K=W_{ext}$$$$=-\left(\mu mg\cos 25°\right)s$$$$=-\frac{1}{4}\times 2.2\times 10\times 2\times \cos 25°$$$$=(-11\cos 25°)J.$$$$\approx -10J.$$

Commented by Tinkutara last updated on 25/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

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