Physics · Friction

JEE Main 2024 — 29 January, Shift 1 — Question 37

A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4 , then the work done against friction (in J ) is :

Option A:
4200
Option B:
3900
Option C:
4000
Correct
Option D:
4500

Answer: C

Step-by-step solution

Given $\mathrm{m}=100 \mathrm{~kg}$

$\mathrm{s}=10 \mathrm{~m}$

$\mu=0.4$ As $\mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N}$

Now $\mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Friction
Topic: Single Block Problems Involving Friction