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

The forces on the block are:
  • weight $$mg$$ downward
  • normal reaction $$N$$ upward
  • applied force $$\mathbf{F}$$ making $$45^{\circ}$$ with the horizontal
  • kinetic friction $$f_k$$ opposite to the motion along the surface.

Resolve the applied force:
  Horizontal component  $$F_x = F\cos 45^{\circ} = \dfrac{F}{\sqrt{2}}$$
  Vertical component    $$F_y = F\sin 45^{\circ} = \dfrac{F}{\sqrt{2}}$$

Vertical equilibrium (no vertical motion):
$$N + F_y = mg$$
so
$$N = mg - \frac{F}{\sqrt{2}} \quad -(1)$$

Friction coefficient is $$\mu = 0.25$$, hence
$$f_k = \mu N = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(2)$$

The block is pulled steadily (horizontal acceleration zero). Therefore
horizontal forces balance:
$$F_x = f_k$$
$$\frac{F}{\sqrt{2}} = \mu\left( mg - \frac{F}{\sqrt{2}} \right) \quad -(3)$$

Solve equation $$-(3)$$ for $$F$$:
$$\frac{F}{\sqrt{2}} + \mu\frac{F}{\sqrt{2}} = \mu mg$$
$$F(1+\mu)\frac{1}{\sqrt{2}} = \mu mg$$
$$F = \frac{\mu mg\sqrt{2}}{1+\mu}$$

Insert the data $$m = 25\ \text{kg},\; g = 9.8\ \text{m s}^{-2},\; \mu = 0.25$$:
$$F = \frac{0.25 \times 25 \times 9.8 \times 1.414}{1 + 0.25}$$
$$F \approx \frac{86.6}{1.25} \approx 69.3\ \text{N}$$

Horizontal component of the applied force:
$$F_x = \frac{F}{\sqrt{2}} \approx \frac{69.3}{1.414} \approx 49.0\ \text{N}$$

Work done by the external force over the displacement $$s = 5\ \text{m}$$:
$$W = F_x\,s = 49.0 \times 5 \approx 245\ \text{J}$$

Therefore, the work done by the external force is $$245\ \text{J}$$.
Correct option: Option C