Question

The potential energy of a particle of mass m free to move along x-axis is given by $U=\frac{1}{2}k{x}^2$ for $x<0$ and $U=0$ for $x\ge 0$ (x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at $x=-\sqrt{\frac{2E}{k}}$ is :

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

Answer 1

The correct option is A $zero$

Given $U=\frac{1}{2}k{x}^2$ (Potential energy)

$E=$Total mechanical energy

$U=\frac{1}{2}k{x}^2$ as $x<0$

$U=0$ for $x\ge 0$

at $x=-\sqrt{\frac{2E}{K}}$ we have to find kinetic energy $(K.E)

as we know from conservation of mechanical enegy

$E=U+K.E$

$U_{x=-\sqrt{2E/K}}=\frac{1}{2}\times K\times \frac{2E}{K}=E$

$K.E_{x=-\sqrt{2E/K}}=E-U=E-E=0$

$\Rightarrow \frac{1}{2}m{v}^2=0$

$\Rightarrow v=0$

Hence speed at $x=-\sqrt{\frac{2E}{k}}$ is zero.