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 determine the work done on the block, we need to consider the forces acting on it and the displacement. The force is applied at an angle of \(45^\circ\) to the horizontal, and there is friction between the block and the surface.

1. The force \( F \) can be divided into two components:
   • Horizontal component: \( F_{\text{horizontal}} = F \cos 45^\circ \)
   • Vertical component: \( F_{\text{vertical}} = F \sin 45^\circ \)
2. Only the horizontal component does work on the block, as work is calculated along the direction of displacement.
3. The maximum frictional force \( F_{\text{friction}} \) is given by: \(F_{\text{friction}} = \mu N = \mu (mg - F_{\text{vertical}})\) where:
   • \( \mu = 0.25 \) (coefficient of friction)
   • \( N = mg - F \sin 45^\circ \) is the normal force
   • \( m = 25 \, \text{kg} \)
   • \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
4. Let's calculate the work done: \(N = mg - F \sin 45^\circ = 25 \times 9.8 - F \times \frac{\sqrt{2}}{2}\)
5. Substitute the values:
   • \(N = 245 - F \times 0.707\)
6. Frictional Force: \(F_{\text{friction}} = 0.25 \times (245 - F \times 0.707)\)
7. The work done against friction is given by: \(W = F_{\text{horizontal}} \times d - F_{\text{friction}} \times d\) where \( d = 5 \, \text{m} \) is the displacement.
8. Assuming \( F_{\text{horizontal}} \) equals the force of friction (slipping condition): \(F \cos 45^\circ = F_{\text{friction}}\)
9. Thus the work done against friction: \(W = F \cos 45^\circ \times 5 - F_{\text{friction}} \times 5 = 245 \, \text{J}\)

Therefore, the correct answer is 245 J.

Answer 2

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).