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| ### Statement |
| ### Statement |
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| $2.3.33.$ [Insert the problem statement] |
| $2.3.33.$ The body slides along a flat surface that smoothly passes into another flat surface located at an angle $\alpha$ to the first one. Coefficient of friction $\mu$. Determine the kinetic energy at the end of the interface section if it is equal to $K_0$ at the beginning. |
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| ### Solution |
| ### Solution |
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| Assume that the speed $v$ of the body is much larger than $\sqrt(gr)$, where $r$ the radius of curvature of the bend, so that the gravitational field $g$ has negligible effect on centripetal acceleration. Consequently, the normal force |
| Assume that the speed $v$ of the body is much larger than $\sqrt{gr}$, where $r$ the radius of curvature of the bend, so that the gravitational field $g$ has negligible effect on centripetal acceleration. Consequently, the normal force |
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| \[N=\frac{mv^2}{r}=\frac{2K}{r},\] |
| \[N=\frac{mv^2}{r}=\frac{2K}{r},\] |
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| where $K$ is the kinetic energy of the body. The change in kinetic energy is due to the work done by friction, i.e. |
| where $K$ is the kinetic energy of the body. The change in kinetic energy is due to the work done by friction, i.e. |
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| \[dK=-\mu Nr\,d\theta=-2\mu K\,d\theta,\] |
| \[dK=-\mu Nr\,d\theta=-2\mu K\,d\theta,\] |
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| where $d\theta$ is a small angle subtended by a small arc of the bend. Integrating, we have $K(\theta)=K_0e^{-2\mu\theta}$. In the figure, the angle subtended by the bend is $\alpha$, so $K=K_0e^{-2\mu\alpha}$ at the end of the bend. |
| where $d\theta$ is a small angle subtended by a small arc of the bend. Integrating, we have $K(\theta)=K_0\cdot e^{-2\mu\theta}$. In the figure, the angle subtended by the bend is $\alpha$, so $K=K_0\cdot e^{-2\mu\alpha}$ at the end of the bend. |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $K_0\cdot e^{-2\mu\alpha}$ |
