Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com