Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

A body of mass $4 kg$ is at rest on a horizontal table. The coefficient of friction between body and table is 0.3 . Find the acceleration of the body if a horizontal force applied on the body is 20 N

\[(a)\;2.06 m/s^2 \quad (b)\;2.6 m/s^2 \quad (c)\;1.82 m/s^2 \quad(d)\;1.06 m/s^2 \]
Can you answer this question?

1 Answer

0 votes
$\quad=4 \times 9.8$
$\quad=39.2 N$
Therefore $ f=\mu mg$
$\quad \qquad= 0.3 \times 39.2N$
$\quad\qquad=11.76 N$
$F-f\; =ma$
$\qquad \quad 8.24=4a$
$\qquad\qquad a= 2.06 m/s^2$
Hence a is the correct answer.


answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App