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प्रश्न
A block of mass 30 kg is pulled up a slope (diagram alongside) with a constant speed by applying a force of 200 N parallel to the slope. A and B are the initial and final positions of the block. Calculate the force of friction offered by the surface AB.
संख्यात्मक
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उत्तर
The block moves at constant speed, so the net force along the slope is zero.
From the diagram, \[ \sin\theta = \frac{\text{height}}{\text{slope length}} \]
\[= \frac{1.5}{3} \]
= 0.5
The component of the block’s weight acting down the slope = mg sin θ
= 30 × 10 × 0.5
= 150 N
The applied force is 200 N up the slope.
Therefore, friction acts down the slope and balances the remaining force.
Friction = 200 − 150 = 50 N
The force of friction is 50 N, acting down the slope.
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