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Question Number 14999 by tawa tawa last updated on 06/Jun/17

$$\mathrm{A}20\mathrm{kg}\mathrm{box}\mathrm{is}\mathrm{released}\mathrm{from}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{an}\mathrm{inclined}\mathrm{plane}\mathrm{that}\mathrm{is}5\mathrm{m}\mathrm{long}\mathrm{and}$$$$\mathrm{makes}\mathrm{an}\mathrm{angle}\mathrm{of}20°\mathrm{to}\mathrm{the}\mathrm{horizontal}.\mathrm{A}60\mathrm{N}\mathrm{friction}\mathrm{force}\mathrm{impedes}\mathrm{the}\mathrm{motion}\mathrm{of}\mathrm{the}$$$$\mathrm{box}.\mathrm{How}\mathrm{long}\mathrm{will}\mathrm{it}\mathrm{take}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{bottom}\mathrm{of}\mathrm{the}\mathrm{box}.$$

Answered by sandy_suhendra last updated on 06/Jun/17

Commented by sandy_suhendra last updated on 06/Jun/17

$$\mathrm{\Sigma }\mathrm{F}=\mathrm{m}.\mathrm{a}$$$$\mathrm{m}.\mathrm{g}\sin 20°-\mathrm{f}=\mathrm{m}.\mathrm{a}$$$$20\times 9.8\times 0.342-60=20\mathrm{a}$$$$7=20\mathrm{a}$$$$\mathrm{a}=0.35\mathrm{m}/{\mathrm{s}}^2$$$$$$$${\mathrm{Vt}}^2={\mathrm{Vo}}^2+2\mathrm{a}.\mathrm{S}$$$${\mathrm{Vt}}^2=0+2\times 0.35\times 5$$$${\mathrm{Vt}}^2=3.5$$$$\mathrm{Vt}=\sqrt{3.5}=1.87\mathrm{m}/\mathrm{s}$$$$$$$$\mathrm{Vt}=\mathrm{Vo}+\mathrm{a}.\mathrm{t}$$$$1.87=0+0.35\mathrm{t}$$$$\mathrm{t}=5.34\mathrm{s}$$

Commented by tawa tawa last updated on 09/Jun/17

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

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