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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2024
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(C) \(4000\)

Step-by-step Solution

Detailed explanation

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 \(W=f . s=400 \times 10=4000 \mathrm{~J}\)

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