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A-body-of-mass-0-3-kg-is-taken-up-an-inclined-plane-of-length-10-m-and-height-5-m-and-then-allowed-to-slide-down-to-the-bottom-again-The-coefficient-of-friction-between-the-body-and-the-plane-is-0-1




Question Number 22884 by Tinkutara last updated on 23/Oct/17
A body of mass 0.3 kg is taken up an  inclined plane of length 10 m and  height 5 m, and then allowed to slide  down to the bottom again. The  coefficient of friction between the body  and the plane is 0.15. What is the  kinetic energy of the body at the end of  the trip?
$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{0}.\mathrm{3}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{taken}\:\mathrm{up}\:\mathrm{an} \\ $$$$\mathrm{inclined}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{length}\:\mathrm{10}\:\mathrm{m}\:\mathrm{and} \\ $$$$\mathrm{height}\:\mathrm{5}\:\mathrm{m},\:\mathrm{and}\:\mathrm{then}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{slide} \\ $$$$\mathrm{down}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{again}.\:\mathrm{The} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{body} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{0}.\mathrm{15}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{trip}? \\ $$
Answered by ajfour last updated on 23/Oct/17
K=mgh−(μmgcos θ)l      =mg(h−μ(√(l^2 −h^2 )) )      =3(5−0.15×5×1.732)J      =(15−2.25×1.732)J      =(15−(9/4)×1.732)J       =(15−9×0.433)J      =(15−3.6−0.3)J = 11.1J
$${K}={mgh}−\left(\mu{mg}\mathrm{cos}\:\theta\right){l} \\ $$$$\:\:\:\:={mg}\left({h}−\mu\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:=\mathrm{3}\left(\mathrm{5}−\mathrm{0}.\mathrm{15}×\mathrm{5}×\mathrm{1}.\mathrm{732}\right){J} \\ $$$$\:\:\:\:=\left(\mathrm{15}−\mathrm{2}.\mathrm{25}×\mathrm{1}.\mathrm{732}\right){J} \\ $$$$\:\:\:\:=\left(\mathrm{15}−\frac{\mathrm{9}}{\mathrm{4}}×\mathrm{1}.\mathrm{732}\right){J}\: \\ $$$$\:\:\:\:=\left(\mathrm{15}−\mathrm{9}×\mathrm{0}.\mathrm{433}\right){J} \\ $$$$\:\:\:\:=\left(\mathrm{15}−\mathrm{3}.\mathrm{6}−\mathrm{0}.\mathrm{3}\right){J}\:=\:\mathrm{11}.\mathrm{1}{J}\: \\ $$
Commented by Tinkutara last updated on 24/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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