Exam

JEE Main

Year

2025

Date

4 April

Shift

Evening Shift

Subject

Physics

Chapter

Work, Power and Energy

JEE Main 2025 · 4 April · Evening Shift · Physics · Work, Power and Energy

A block of mass 25 kg is pulled along a horizontal surface by a...

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:

Step-by-step solution

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