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Question Number 22884 by Tinkutara last updated on 23/Oct/17

$$\mathrm{A}\mathrm{body}\mathrm{of}\mathrm{mass}0.3\mathrm{kg}\mathrm{is}\mathrm{taken}\mathrm{up}\mathrm{an}$$$$\mathrm{inclined}\mathrm{plane}\mathrm{of}\mathrm{length}10\mathrm{m}\mathrm{and}$$$$\mathrm{height}5\mathrm{m},\mathrm{and}\mathrm{then}\mathrm{allowed}\mathrm{to}\mathrm{slide}$$$$\mathrm{down}\mathrm{to}\mathrm{the}\mathrm{bottom}\mathrm{again}.\mathrm{The}$$$$\mathrm{coefficient}\mathrm{of}\mathrm{friction}\mathrm{between}\mathrm{the}\mathrm{body}$$$$\mathrm{and}\mathrm{the}\mathrm{plane}\mathrm{is}0.15.\mathrm{What}\mathrm{is}\mathrm{the}$$$$\mathrm{kinetic}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{of}$$$$\mathrm{the}\mathrm{trip}?$$

Answered by ajfour last updated on 23/Oct/17

$$K=mgh-\left(\mu mg\cos \theta \right)l$$$$=mg(h-\mu \sqrt{l^2-h^2})$$$$=3(5-0.15\times 5\times 1.732)J$$$$=(15-2.25\times 1.732)J$$$$=(15-\frac{9}{4}\times 1.732)J$$$$=(15-9\times 0.433)J$$$$=(15-3.6-0.3)J=11.1J$$

Commented by Tinkutara last updated on 24/Oct/17

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

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