Question

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $ 45^\circ $ with the horizontal. The friction coefficient between the block and the surface is 0.25. The displacement of 5 m of the block is:

• A. 970 J
• B. 735 J
• C. 245 J
• D. 490 J

Hint:

When a block moves with uniform velocity, the work done by the applied force equals the work done against friction.

Answer 1

The Correct Option is C

Given: - \( m = 25 \, \text{kg} \), 
- The force \( F \) is applied at an angle \( 45^\circ \) with the horizontal, 
- The coefficient of friction \( \mu = 0.25 \), 
- The displacement \( d = 5 \, \text{m} \). 
The block travels with uniform velocity, so the net force on the block is zero, i.e., the force applied equals the frictional force. 
The frictional force \( F_f \) is given by: \[ F_f = \mu mg = 0.25 \times 25 \times 9.8 = 61.25 \, \text{N} \] The work done by the frictional force is: \[ W_f = F_f \times d = 61.25 \times 5 = 305.25 \, \text{J} \] Now, the force \( F \) applied at an angle is: \[ F = \frac{61.25}{\cos 45^\circ} = \frac{61.25}{\frac{1}{\sqrt{2}}} = 61.25 \times \sqrt{2} = 86.5 \, \text{N} \] The work done by the applied force is: \[ W_{\text{applied}} = F \times d = 86.5 \times 5 = 432.5 \, \text{J} \] 
Thus, the work done by the applied force is 432.5 J, and the correct answer is (3).