Question

A massless spring of natural length of $0.5m$ and spring constant $50N/m$ has one end fixed and the other end attached to a mass of $250g$. The spring mass system is on a smooth floor. The mass is pulled until the length of the spring is $0.6$ and then released from rest. The kinetic energy of the mass when the length of the spring is $0.5m$ is

• A. $0.25J$
• B. $2.25J$
• C. $6.25J$
• D. $9J$

Answer 1

The correct option is A $0.25J$

Kinetic energy of mass at equilibrium position $=$ potential energy of spring at max stretch

So, $KE=\frac{1}{2}k{x}^2$

or $KE=\frac{1}{2}\times 50\times (0.1{)}^2$

or $KE=0.25J$