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A-20-kg-box-is-released-from-the-top-of-an-inclined-plane-that-is-5-m-long-and-makes-an-angle-of-20-to-the-horizontal-A-60N-friction-force-impedes-the-motion-of-the-box-How-long-will-it-take-to-re




Question Number 14999 by tawa tawa last updated on 06/Jun/17
A 20 kg box is released from the top of an inclined plane that is 5 m long and  makes an angle of 20° to the horizontal. A 60N friction force impedes the motion of the  box . How long will it take to reach the bottom of the box.
$$\mathrm{A}\:\mathrm{20}\:\mathrm{kg}\:\mathrm{box}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{is}\:\mathrm{5}\:\mathrm{m}\:\mathrm{long}\:\mathrm{and} \\ $$$$\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{20}°\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{A}\:\mathrm{60N}\:\mathrm{friction}\:\mathrm{force}\:\mathrm{impedes}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{box}\:.\:\mathrm{How}\:\mathrm{long}\:\mathrm{will}\:\mathrm{it}\:\mathrm{take}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$
Answered by sandy_suhendra last updated on 06/Jun/17
Commented by sandy_suhendra last updated on 06/Jun/17
ΣF=m.a  m.g sin 20°−f = m.a  20×9.8×0.342−60=20a  7=20a  a=0.35 m/s^2     Vt^2 =Vo^2 +2a.S  Vt^2 =0+2×0.35×5  Vt^2 =3.5  Vt=(√(3.5)) = 1.87 m/s    Vt = Vo + a.t  1.87=0+0.35t  t=5.34 s
$$\Sigma\mathrm{F}=\mathrm{m}.\mathrm{a} \\ $$$$\mathrm{m}.\mathrm{g}\:\mathrm{sin}\:\mathrm{20}°−\mathrm{f}\:=\:\mathrm{m}.\mathrm{a} \\ $$$$\mathrm{20}×\mathrm{9}.\mathrm{8}×\mathrm{0}.\mathrm{342}−\mathrm{60}=\mathrm{20a} \\ $$$$\mathrm{7}=\mathrm{20a} \\ $$$$\mathrm{a}=\mathrm{0}.\mathrm{35}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Vt}^{\mathrm{2}} =\mathrm{Vo}^{\mathrm{2}} +\mathrm{2a}.\mathrm{S} \\ $$$$\mathrm{Vt}^{\mathrm{2}} =\mathrm{0}+\mathrm{2}×\mathrm{0}.\mathrm{35}×\mathrm{5} \\ $$$$\mathrm{Vt}^{\mathrm{2}} =\mathrm{3}.\mathrm{5} \\ $$$$\mathrm{Vt}=\sqrt{\mathrm{3}.\mathrm{5}}\:=\:\mathrm{1}.\mathrm{87}\:\mathrm{m}/\mathrm{s} \\ $$$$ \\ $$$$\mathrm{Vt}\:=\:\mathrm{Vo}\:+\:\mathrm{a}.\mathrm{t} \\ $$$$\mathrm{1}.\mathrm{87}=\mathrm{0}+\mathrm{0}.\mathrm{35t} \\ $$$$\mathrm{t}=\mathrm{5}.\mathrm{34}\:\mathrm{s} \\ $$
Commented by tawa tawa last updated on 09/Jun/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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