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 performed by a force is calculated as the dot product of the force vector and the displacement vector: \( W = \vec{F} \cdot \vec{d} = F d \cos\theta \). Given values are: - \( F = 20 \, \text{N} \) (magnitude of the applied force), - \( d = 10 \, \text{m} \) (displacement), - \( \theta = 30^\circ \) (the angle between the applied force and the horizontal displacement). The horizontal component of the applied force is: \[ F_x = F \cos\theta = 20 \cos 30^\circ \] Since \( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \), the horizontal force component is: \[ F_x = 20 \times 0.866 \approx 17.32 \, \text{N} \] The work done by the applied force is then: \[ W = F_x \cdot d = 17.32 \times 10 = 173.2 \, \text{J} \] Alternatively, applying the direct formula: \[ W = F d \cos\theta = 20 \times 10 \times \cos 30^\circ = 200 \times 0.866 \approx 173.2 \, \text{J} \] The work done by the applied force is \( 173.2 \, \text{J} \). Frictional or normal forces are not relevant for this specific calculation of work done by the applied force.