Question

The potential energy of a particle of mass $m$ is given by $U=\frac{1}{2} k x^{2}$ for $x<0$ and $U=0$ for $x \geq 0$. If total mechanical energy of the particle is $E$. Then its speed at $x=\sqrt{\frac{2 E}{k}}$ is :

• A. Zero
• B. $\sqrt{\frac{2 E}{m}}$
• C. $\sqrt{\frac{E}{m}}$
• D. $\sqrt{\frac{E}{2 m}}$

Answer 1

Answer: (B)

$U =\frac{1}{2} k x^{2}, x< 0 $
$ U =0, x \geq 0 $
$K+U =E $
$\frac{1}{2} m v^{2} =E-U $
For $x \geq 0$,
$U= 0$
[As $.x=\sqrt{\frac{2 E}{K}}$]
$\Rightarrow \frac{1}{2} m v^{2}=E$
$v=\sqrt{\frac{2 E}{m}}$