Question

A spring mass system is given on which a force F is applied towards right side. Due to which the spin stretches to a distance x . Derive an expression for the potential energy stored in the spring.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Potential\:energy=\frac{1}{2}kx^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Force \: applied = F \\ \\ \tt: \implies Stretched \: length = x \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Potential \: energy \: in \: spring =?$

• According to given question :

• Force applied is balanced by spring constant.

$\bold{As \: we \: know \: that} \\ \tt: \implies F = kx - - - - - (1)$

• Let dw be work done for a small interval of time with dx displacement in the spring.

$\bold{As \: we \: know \: that} \\ \tt: \implies dw = F \: dx \\ \\ \tt \circ \: Putting \: value \: of \: (1) \\ \tt: \implies dw = kx \: dx \\ \\ \tt: \implies \int dw = \int kx \: dx \\ \\ \tt: \implies \int \limits_{0}^{w} dw = k \int \limits_{0}^{x} x \: dx \\ \\ \tt: \implies w = k \bigg( \frac{ {x}^{2} }{2} \bigg) \\ \\ \green{ \tt: \implies w = \frac{1}{2} k {x}^{2} }$

• Here work done by spring is potential energy conserved in the spring.

$\green{ \tt \therefore Potential \: energy \: in \: spring \: is \: \frac{1}{2} {kx}^{2} }$