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

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive x-axis direction is applied to the block . The force $\hat F$ is given by $\hat F=(4-x^2)\hat i \;N$ . Initial position of the block is $x=0$. The maximum kinetic energy of block between $x=0$ to $x=2.0$ is

\[(a)\;5.33\;J \quad (b)\;8.67\;J \quad (c)\; 2.33 \;J \quad (d)\;6.67\;J\]

Can you answer this question?

1 Answer

0 votes
From work energy theorem
Kinetic energy of block
$k=\int \limits_0^x (4-x^2)dx$
For K to be maximum $\large\frac{dh}{dx}$$=0$
$x=\pm 2$
At $x=+2 \quad \large\frac{d^2k}{dx^2}$ is -ve . So k is maximum
answered Aug 6, 2013 by meena.p
edited Jun 6, 2014 by lmohan717

Related questions

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