Question

A spring block arrangement is shown in the figure.


The spring is stretched along the inclined plane from position $A$ (unstretched) to position $B$. Find the change in potential energy of the spring.

• A. $\frac{1}{2}k(x\sin \theta {)}^2+mgx\sin \theta$
• B. $\frac{1}{2}k(x\cos \theta {)}^2$
• C. $\frac{1}{2}k{x}^2\sin \theta -mgx\sin \theta$
• D. $\frac{1}{2}k{x}^2$

Answer 1

The correct option is D $\frac{1}{2}k{x}^2$


If the spring is stretched by distance $x$,
then, change in potential energy of spring, $U=\frac{1}{2}k{x}^2$
Note: Spring potential energy is conservative in nature. So, stretched length does not depend upon inclination of surface.