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  • A block mass m slides along a floor while a force of magnitude F is applied to it at an angle as shown in figure. The coefficient of kinetic friction is <img alt="\mu _{K}." src="/latex-image/?%5Cmu%20_

A block mass m slides along a floor while a force of magnitude F is applied to it at an angle \theta as shown in figure. The coefficient of kinetic friction is \mu _{K}. Then the block's acceleration 'a' is given by : (g is acceleration due to gravity)
 
Option: 1 -\frac{F}{m}\cos \theta -\mu _{K}\left ( g-\frac{F}{m}\sin \theta \right )
Option: 2 \frac{F}{m}\cos \theta -\mu _{K}\left ( g-\frac{F}{m}\sin \theta \right )
Option: 3 \frac{F}{m}\cos \theta +\mu _{K}\left ( g-\frac{F}{m}\sin \theta \right )  
Option: 4 \frac{F}{m}\cos \theta -\mu _{K}\left ( g+\frac{F}{m}\sin \theta \right )

Answers (1)

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\begin{array}{l} \mathrm{N}=\mathrm{mg}-\mathrm{f} \sin \theta ....(1)\\ \\ \mathrm{F} \cos \theta-\mu_{\mathrm{k}} \mathrm{N}=\mathrm{ma} ... \\ \\ Using \ equation \ (1) \\\\ \mathrm{F} \cos \theta-\mu_{\mathrm{k}}(\mathrm{mg}-\mathrm{F} \sin \theta)=\mathrm{ma} \\ \\ \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right) \end{array}

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