Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is

  • Option 1)

    0.02

  • Option 2)

    0.03

  • Option 3)

    0.06

  • Option 4)

    0.01

 

Answers (1)

As we learnt in

Kinetic or Dynamic Friction -

f_{K}\;\alpha\ R

f_{K}=\mu_{K} R

f_{K}= kinetic friction 

\mu_{K}= coefficient of kinetic friction

R = reaction

- wherein

f_{K}<F_{l}

\therefore\ \mu_{K}<\mu_{s}

\mu_{K}=depends on the nature of surface in contact.

 

 

 

: Frictional force provides the retarding force

\therefore\: \: \: \mu mg=ma

or     \mu=\frac{a}{g}= \frac{u/t}{g}= \frac{6/10}{10}= 0.06

Correct option is 3.


Option 1)

0.02

This is an incorrect option.

Option 2)

0.03

This is an incorrect option.

Option 3)

0.06

This is the correct option.

Option 4)

0.01

This is an incorrect option.

Posted by

Sabhrant Ambastha

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

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question