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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2025
  • Medium
  • MCQ

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:

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Answer & Solution

Correct Answer

(C) 245 J

Step-by-step Solution

Detailed explanation


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