JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 13)
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 block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is :
490 J
970 J
735 J
245 J
Explanation
_4th_April_Evening_Shift_en_13_1.png)
Block travels with uniform velocity
So $\quad \mathrm{a}=0 \Rightarrow \mathrm{~F} \cos 45^{\circ}=$ friction
$$\begin{aligned} & \frac{\mathrm{F}}{\sqrt{2}}=\mu\left[\mathrm{mg}-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \frac{\mathrm{F}}{\sqrt{2}}=0.25\left[25 \times 9.8-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \Rightarrow \quad 1.25 \frac{\mathrm{~F}}{\sqrt{2}}=61.25 \\ & \mathrm{~F}=\frac{61.25 \times \sqrt{2}}{1.25}=49 \sqrt{2} \\ & \mathrm{~W}_{\mathrm{ext}}=\mathrm{FS} \cos 45^{\circ} \\ &=49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}}=245 \mathrm{~J} \end{aligned}$$
Comments (0)
