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Question Number 178609 by Tawa11 last updated on 18/Oct/22

Answered by mr W last updated on 19/Oct/22

$$\left(i\right)$$$$\frac{{mv}_0^2}{2}=mgs\sin \theta +\mu mg\cos \theta s$$$$s=\frac{v_0^2}{2g(\sin \theta +\mu \cos \theta )}$$$$=\frac{{12}^2}{2\times 9.81(\frac{1}{2}+\frac{0.16\times \sqrt{3}}{2})}\approx 11.49m$$$$\left(ii\right)$$$$\frac{m(v_0^2-v_1^2)}{2}=2s\times \mu mg\cos \theta$$$$\frac{m(v_0^2-v_1^2)}{2}=2\times \frac{v_0^2}{2g(\sin \theta +\mu \cos \theta )}\times \mu mg\cos \theta$$$$v_1=v_0\sqrt{1-\frac{2\mu \cos \theta }{\sin \theta +\mu \cos \theta }}$$$$\Rightarrow v_1=v_0\sqrt{\frac{\sin \theta -\mu \cos \theta }{\sin \theta +\mu \cos \theta }}$$$$=12\sqrt{\frac{1-0.16\sqrt{3}}{1+0.16\sqrt{3}}}\approx 9.03m/s$$

Commented by Tawa11 last updated on 19/Oct/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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