Question

A block of mass 5 kg is pulled along a horizontal surface by a force of 20 N at an angle of 30° to the horizontal. If the coefficient of friction between the block and the surface is 0.2 and the acceleration due to gravity is \( 10 \, \text{m/s}^2 \), what is the work done by the applied force in moving the block 10 m?

• A. \( 100 \, \text{J} \)
• B. \( 173.2 \, \text{J} \)
• C. \( 200 \, \text{J} \)
• D. \( 346.4 \, \text{J} \)

Hint:

When calculating work done by a specific force, use only the component of the force along the direction of displacement. For forces at an angle, multiply by \( \cos\theta \) to get the effective force component.

Answer 1

The Correct Option is B

The work done by a force is given by: \[ W = \vec{F} \cdot \vec{d} = F d \cos\theta \] Where: - \( F = 20 \, \text{N} \) (magnitude of the applied force), - \( d = 10 \, \text{m} \) (displacement), - \( \theta = 30^\circ \) (angle between the force and the displacement, which is horizontal). The horizontal component of the applied force is: \[ F_x = F \cos\theta = 20 \cos 30^\circ \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \] \[ F_x = 20 \times 0.866 \approx 17.32 \, \text{N} \] The work done by the applied force is: \[ W = F_x \cdot d = 17.32 \times 10 = 173.2 \, \text{J} \] Alternatively, using the formula directly: \[ W = F d \cos\theta = 20 \times 10 \times \cos 30^\circ = 200 \times 0.866 \approx 173.2 \, \text{J} \] Note that the frictional force and normal force do not affect the work done by the applied force, as the question specifically asks for the work done by the 20 N force. The frictional force would be relevant if calculating net work or acceleration, but it is not needed here. Thus, the work done by the applied force is \( 173.2 \, \text{J} \).