Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

The potential energy of a 1 kg mass free to move along the x-axis is given by $V_{(x)}=\bigg[\large\frac{x^4}{4}-\frac{x^2}{2}\bigg]$$J$ The total mechanical energy of the particle is $2J$ then the maximum speed is?

\[(a)\;\frac{3}{\sqrt 2}\; m/s \quad (b)\;\sqrt 2 \;m/s\quad (c)\; \frac{1}{\sqrt 2}\; m/s \quad (d)\;2\; m/s\]

Can you answer this question?

1 Answer

0 votes
Velocity is maximum when $KE$ of particle is maximum. ie potential energy of particle is minimum
$\large\frac{dV}{dx}$$=0=>x^3-x=0=>x=\pm 1$
minimum $PE$ is when $x=1$
min $PE=\large\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}$$J$
Given $KE_{\large \max}$$+PE_{\large min}$$=2\;J$
$KE_{\large max }=2 +\large\frac{1}{4}=\frac{9}{4}$
$\large\frac{9}{4}=\frac{1}{2} $$mv^2_{\large max} \qquad (m=1 kg)$
$ v^2 _{\large max}=\large\frac{9}{2}$
$v_{\large max}=\large\frac{3}{\sqrt 2}$$\;m/s$
Hence a is the correct answer.


answered Aug 6, 2013 by meena.p
edited Feb 17, 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