Question

An object is attached horizontally on a frictionless surface to a given spring which obeys Hooke's Law. When the object is at some position $A$, the force felt by the object from the spring is $F=-2.4N$ and the potential energy in the spring is $U=2.4J$. At some different position $B$, the force felt by the object is $F=-12N$.
What is the potential energy stored in the spring when the object is at position $B$?

• A. $12J$
• B. $50J$
• C. $60J$
• D. $100J$
• E. $120J$

Answer 1

The correct option is C $60J$

Force acting at the end of spring compressed or stretched by $x$ is

$F=\left|kx\right|$

$⟹F_1=2.4=k{x}_1$

and $F_2=12=k{x}_2$

$⟹x_2=5x_1$
Potential energy stored is

$U_1=\frac{1}{2}k{x}_1^2=2.4J$
Hence $U_2=\frac{1}{2}k{x}_2^2=60J$