Question

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.

Answer 1

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.