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

To calculate the work done on the block, consider the forces acting on it and its displacement. The applied force is at \(45^\circ\) to the horizontal, and friction is present.

1. The applied force \( F \) has components:
   • Horizontal: \( F_{\text{horizontal}} = F \cos 45^\circ \)
   • Vertical: \( F_{\text{vertical}} = F \sin 45^\circ \)
2. Work is done only by the horizontal component of the force, as displacement is horizontal.
3. The maximum frictional force \( F_{\text{friction}} \) is calculated as: \(F_{\text{friction}} = \mu N = \mu (mg - F_{\text{vertical}})\) with:
   • \( \mu = 0.25 \) (coefficient of friction)
   • \( N = mg - F \sin 45^\circ \) (normal force)
   • \( m = 25 \, \text{kg} \)
   • \( g = 9.8 \, \text{m/s}^2 \)
4. Calculating the normal force: \(N = mg - F \sin 45^\circ = 25 \times 9.8 - F \times \frac{\sqrt{2}}{2}\)
5. Substituting values yields:
   • \(N = 245 - F \times 0.707\)
6. The frictional force is: \(F_{\text{friction}} = 0.25 \times (245 - F \times 0.707)\)
7. Work done against friction is: \(W = (F_{\text{horizontal}} - F_{\text{friction}}) \times d\) where \( d = 5 \, \text{m} \).
8. Assuming the horizontal force equals the frictional force at the point of slipping: \(F \cos 45^\circ = F_{\text{friction}}\)
9. Therefore, the work done against friction is: \(W = F \cos 45^\circ \times 5 - F_{\text{friction}} \times 5 = 245 \, \text{J}\)

The computed work done is 245 J.