Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A body of mass $10 \mathrm{~kg}$ is moving with an initial speed of $20 \mathrm{~m} / \mathrm{s}$ . The body stops after $5 \mathrm{~s}$ due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity $.g=10 \mathrm{~ms}^{-2}$ )
Option: 1
0.3

Option: 2
0.5

Option: 3
0.2

Option: 4
0.4

Answers (1)

best_answer

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$0=20-\mu g(5)$
$\mu=\frac{2}{5}=0.4$

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates

anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'

anam-khan

BTech Admissions 2025 Live: JEE Mains, SRMJEE, VITEEE dates out; AEEE, MET registrations underway

Latest Question