JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 :
- A \(4200\)
- B \(3900\)
- C \(4000\)
- D \(4500\)
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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