Physics · Work, Power & Energy

JEE Main 2025 — 4 April, Evening Shift — Question 59

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 :

Option A:
735 J
Option B:
490 J
Option C:
970 J
Option D:
245 J
Correct

Answer: D

Step-by-step solution

$f=\frac{F}{\sqrt{2}}$

$\implies \mu\left(m g-\frac{F}{\sqrt{2}}\right)=\frac{F}{\sqrt{2}}$

$\implies \frac{1}{4}\left(245-\frac{F}{\sqrt{2}}\right)=\frac{F}{\sqrt{2}}$

$245=5 \frac{F}{\sqrt{2}}$

$\implies \frac{F}{\sqrt{2}}=45$

$W=\frac{F}{\sqrt{2}} \times 5$

$=245 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Work Done by a Constant and Variable Force