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JAMB Physics 2012 Paper

Question : 2

Total: 42

Solution:

Concept:

At uniform speed, the net horizontal force is zero, so friction equals the horizontal pull.

Explanation:

The rope makes an angle of

45^\circ

with the floor.

The horizontal component of the pulling force is:

F_{\text{horizontal}} = 1500 \cos 45^\circ

Since

\cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071

F_{\text{horizontal}} = 1500 \times 0.7071 \approx 1060.65 \, \text{N}

Because the load moves at uniform speed, the frictional force equals this horizontal component.

So, the frictional force is approximately

1061 \, \text{N}

Answer:

1061 \, \text{N}

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