A body moves a distance of $10 ext{ m}$ along a straight line under the action of a force of $5 ext{ N}$.

Question

A body moves a distance of $10 	ext{ m}$ along a straight line under the action of a force of $5 	ext{ N}$. If the work done is $25 	ext{ joules}$, the angle which the force makes with the direction of motion of the body is:

Accepted Answer

The work done by a constant force is defined as the product of the component of the force in the direction of the displacement and the magnitude of this displacement. It is given by the formula:
$$W = Fd \cos 	heta$$ 
Where:
$W = 	ext{work done} = 25 	ext{ J}$
$F = 	ext{force} = 5 	ext{ N}$
$d = 	ext{displacement} = 10 	ext{ m}$
$	heta = 	ext{angle between the force and the displacement vectors}$

Substituting the given values into the formula:
$$25 = 5 	imes 10 	imes \cos 	heta$$
$$25 = 50 \cos 	heta$$
$$\cos 	heta = \frac{25}{50} = \frac{1}{2}$$

Since $\cos 60^\circ = \frac{1}{2}$, the angle $	heta$ is $60^\circ$.

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