logo

Ask Questions, Get Answers

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

A particle of mass $0.3 kg$ is subjected to a force $F=-kx$ with $k=15 N/m$. What will be its initial acceleration if it is released from a point $20 cm$ away from orgin?

\[(a)\;3\;m/s^2 \quad (b)\;15\; m/s^2\quad (c)\;5\;m/s^2 \quad (d)\;-10\;m/s^2\]

Can you answer this question?
 
 

1 Answer

0 votes
$F=-kx=ma$
$a=\large\frac{-kx}{m}$
$a=\large\frac{-(15)(20 \times 10 ^{-2})}{0.3}$
$\quad=-10 m/s^2$
Hence d is the correct answer. 

 

answered Jul 15, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

Related questions

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