Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A square carpet of a mass of 20 kg is dragged from one room into another as shown in the figure. The width of the corridor and the carpet is the same, i.e., 2 m. The friction coefficient is 0.1 at the end of the first room and 0.2 at the other end, in the corridor it changes linearly from 0.1 to 0.2. The carpet presses the floor uniformly. Find the amount of work (in J) that must be done to drag the carpet slowly from the first room to the other.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 60.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Let the coefficient of friction for the corridor is given asμ(x)=0.1+(0.2−0.1)x2Let us divide the carpet into infinitesimally thin vertical strips. The mass per unit width of the carpet beλ=20 kg2 m=10 kg mWork done in moving each strip (dl) through a distance dx is μ(λdl)gdx. Total work done in moving each strip from one end to the other=∫02 0.1+0.12xλdlgdx=(0.3)λdlgTotal work done in moving the carpet =∫02 0.3λgdl⇒0.6×λg=60 J
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring