A block of mass m slides along a floor while a force of magnitude F

Question:

A block of 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 $\mu_{\mathrm{K}}$. Then, the block's acceleration ' $\mathrm{a}$ ' is given by : ( $\mathrm{g}$ is acceleration due to gravity)

  1. (1) $-\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

  2. (2) $\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

  3. (3) $\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}+\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)^{\prime}$

  4. (4) $\frac{\mathrm{F}}{\mathrm{m}} \cos \theta+\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

Correct Option: , 2

Solution:

$\mathrm{N}=\mathrm{mg}-\mathrm{f} \sin \theta$

$\mathrm{F} \cos \theta-\mu_{\mathrm{k}} \mathrm{N}=\mathrm{ma}$

$\mathrm{F} \cos \theta-\mu_{\mathrm{k}}(\mathrm{mg}-\mathrm{F} \sin \theta)=\mathrm{ma}$

$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$

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