A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is (√3)/2. The force to be applied on the block so that the block will move down without acceleration is N. (g=10 m/s²).

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(2) 12.5. Idea: check whether gravity can drag the block down on its own. Here it cannot — friction is stronger — so the applied force has to push the block down the slope. Gravity along the slope: mg sin 30°=5×10×1/2=25 N, down. Friction, opposing the downward motion, so acting up the slope: μ mg cos 30°=(√3)/2×5×10×(√3)/2=3/4×50=37.5 N. Friction exceeds gravity's pull by 12.5 N, so the block would not slide at all by itself. For motion down at constant velocity: F+mg sin 30°=μ mg cos 30° F=37.5-25=12.5 N, directed down the incline. The sign of μ-tan θ is the thing to notice: (√3)/2>tan 30°=1/(√3), so this slope is below the angle of repose and the block has to be pushed, not held.

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is (√3)/2. The force to be applied on the block so that the block will move down without acceleration is ______ N. (g=10 m/s²).

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