A-one-kg-ball-rolling-on-a-smooth-horizontal-surface-at-20-m-s-1-comes-to-the-bottom-of-an-inclined-plane-making-an-angle-of-30-with-the-horizontal-Calculate-K-E-of-the-ball-when-it-is-at-the-b

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Question Number 24022 by Tinkutara last updated on 11/Nov/17

$$\mathrm{A}\mathrm{one}\mathrm{kg}\mathrm{ball}\mathrm{rolling}\mathrm{on}\mathrm{a}\mathrm{smooth}$$$$\mathrm{horizontal}\mathrm{surface}\mathrm{at}20\mathrm{m}{\mathrm{s}}^{-1}\mathrm{comes}\mathrm{to}$$$$\mathrm{the}\mathrm{bottom}\mathrm{of}\mathrm{an}\mathrm{inclined}\mathrm{plane}\mathrm{making}$$$$\mathrm{an}\mathrm{angle}\mathrm{of}30°\mathrm{with}\mathrm{the}\mathrm{horizontal}.$$$$\mathrm{Calculate}\mathrm{K}.\mathrm{E}.\mathrm{of}\mathrm{the}\mathrm{ball}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{at}$$$$\mathrm{the}\mathrm{bottom}\mathrm{of}\mathrm{incline}.\mathrm{How}\mathrm{far}\mathrm{up}\mathrm{the}$$$$\mathrm{incline}\mathrm{will}\mathrm{the}\mathrm{ball}\mathrm{roll}?\mathrm{Neglect}$$$$\mathrm{friction}.$$

Answered by mrW1 last updated on 12/Nov/17

$$KE=\frac{1}{2}{mv}^2+\frac{1}{2}\times \frac{2{mr}^2}{5}\times {\left(\frac{v}{r}\right)}^2=\frac{7}{10}\times {mv}^2$$$$=\frac{7}{10}\times 1\times {20}^2=280J$$$$$$$$mgL\sin \theta =KE=\frac{7}{10}{mv}^2$$$$\Rightarrow L=\frac{7v^2}{10g\sin \theta }=\frac{7\times {20}^2}{10\times 10\times \frac{1}{2}}=56m$$

Commented by Tinkutara last updated on 12/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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