Question 5 of 183

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:

A970 JB735 JC245 JD490 J

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Answer: (C) 245 J

Block travels with uniform velocity
So $\quad \mathrm{a}=0 \Rightarrow \mathrm{~F} \cos 45^{\circ}=$ friction
$\begin{aligned}
& \frac{\mathrm{F}}{\sqrt{2}}=\mu\left[\mathrm{mg}-\frac{\mathrm{F}}{\sqrt{2}}\right] \\
& \frac{\mathrm{F}}{\sqrt{2}}=0.25\left[25 \times 9.8-\frac{\mathrm{F}}{\sqrt{2}}\right] \\
& \Rightarrow \quad 1.25 \frac{\mathrm{~F}}{\sqrt{2}}=61.25 \\
& \mathrm{~F}=\frac{61.25 \times \sqrt{2}}{1.25}=49 \sqrt{2} \\
& \mathrm{~W}_{\mathrm{ext}}=\mathrm{FS} \cos 45^{\circ} \\
&=49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}}=245 \mathrm{~J}
\end{aligned}$