A block of mass $25\ \text{kg}$ 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 work done by the force for a displacement of $5\ \text{m}$ of the block (moving with uniform velocity) is: (take $g = 9.8\ \text{m/s}^2$)
Answer: (C) $245\ \text{J}$
The upward component of $F$ reduces the normal force: $N = mg - F\sin45^\circ$.
At uniform velocity the horizontal forces balance:
$$F\cos45^\circ = \mu\left(mg - F\sin45^\circ\right) \Rightarrow \frac{F}{\sqrt2}(1 + 0.25) = 0.25\times25\times9.8 = 61.25$$
$$\frac{F}{\sqrt2} = 49\ \text{N}$$
Work done by $F$: $W = F\cos45^\circ\times5 = 49\times5 = 245\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics