Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2024 · 29 January · Morning Shift · Physics · Work, Power and Energy

A block of mass 100kg slides over a distance of 10m on a horizontal...

A block of mass 100kg100kg slides over a distance of 10m10m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.40 . 4 , then the work done against friction (in JJ) is:

Options

  1. A42004200
  2. B39003900
  3. C40004000
  4. D45004500

Correct answer

C. 40004000

Step-by-step solution

Given m=100kgm = 100kg

s=10ms = 10m

μ=0.4\mu = 0 . 4

The frictional force is given by

 {f=\mumg=0.4×100×10=400N\begin{cases}f & = & \mumg \\ & = & 0 . 4 \times 100 \times 10 \\ & = & 400N\end{cases}

So, the work done by the force can be calculated as follows:

{W=f×s=400×10=4000J\begin{cases}W & = & f \times s \\ & = & 400 \times 10 \\ & = & 4000J\end{cases}

Practice Work, Power and Energy on Quantrex Academy →

More from Work, Power and Energy

All Work, Power and Energy questions Full Work, Power and Energy list All JEE Main PYQs