Question - Calculating Work Done by a Force at an Angle

The work \( W \) done by a force \( \vec{F} \) on an object through a displacement \( \vec{d} \) is given by the dot product of the force and displacement vectors:

\( W = \vec{F} \cdot \vec{d} \)

When the force is applied at an angle \( \theta \) to the direction of displacement, the work done is:

\( W = F \cdot d \cdot \cos(\theta) \)

Given the magnitude of the force \( F = 50 \, \text{N} \), the displacement \( d = 10 \, \text{m} \), and the angle \( \theta = 30^\circ \), we can calculate the work done as follows:

\( W = 50 \, \text{N} \cdot 10 \, \text{m} \cdot \cos(30^\circ) \)

\( W = 500 \, \text{N} \cdot \text{m} \cdot \frac{\sqrt{3}}{2} \)

\( W = 250\sqrt{3} \, \text{J} \)

Thus, the work done is \( 250\sqrt{3} \, \text{Joules} \).