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(D) First, analyze the forces acting on the block in the vertical direction to find the normal force $N$:$N = W + \frac{W}{2} - \frac{W}{2} \sin(30^\c

Vedclass · Accepted Answer

For $\mu < \frac{\sqrt{3}}{5}$, the block will move.

Vedclass · Answer

(D) First, analyze the forces acting on the block in the vertical direction to find the normal force $N$:$N = W + \frac{W}{2} - \frac{W}{2} \sin(30^\circ) = W + \frac{W}{2} - \frac{W}{4} = \frac{5}{4} W$.Next, analyze the forces in the horizontal direction. The driving force is $F_x = \frac{W}{2} \cos(30^\circ) = \frac{W}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} W}{4}$.The limiting friction force is $f_L = \mu N = \mu \cdot \frac{5}{4} W$.The block will move if the driving force is greater than the limiting friction force, i.e., $F_x > f_L$.$\frac{\sqrt{3} W}{4} > \mu \cdot \frac{5}{4} W \Rightarrow \mu Therefore, if $\mu < \frac{\sqrt{3}}{5}$, the block will move. If the block does not move, the work done by static friction is zero. Thus, option $D$ is correct.

$A$ block of weight $W$ is kept on a rough horizontal surface (friction coefficient $\mu$). Two forces of magnitude $W/2$ each are applied as shown in the figure. Choose the $\text{CORRECT}$ statement:

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