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JAMB Physics 2024 Paper

Question : 62

Total: 125

Using the diagram above, the effective force pushing it forward at an angle 60º is

Solution:

Concept:

The effective forward force is the horizontal component of the applied force.

Explanation:

The horizontal component is calculated using

F_x = F \cos \theta

From the diagram, the applied force is

F = 50\, \text{N}

and it acts at an angle

\theta = 60^\circ

to the horizontal.

Substitute the values:

F_x = 50 \cos 60^\circ = 50 \times 0.5 = 25.00\, \text{N}

Thus, the effective force pushing the object forward is

25.00\, \text{N}

Answer:

A. 25.00N

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