Question

derive an expression of potential energy of an elastic stretch of a spring. Prove that the elastic force a spring is conservation ​

Answer 1

Answer:

We can compute Elastic potential energy by using fundamental formula as below:

Elastic potential energy = force

×

displacement.

It is computed as the work done to stretch the spring which depends on the spring constant k and the displacement stretched.

According to Hooke’s law, the force applied to stretch the spring is directly proportional to the amount of stretch. In other words,

The force required to stretch the spring is directly proportional to its displacement. It is given as

P.E. = Magnitude of Force × Displacement

P.E. =

1

2

k

x

2

-ve sign indicates the opposite direction.

Where,

P.E. Elastic Potential Energy

k Spring Constant

x Displacement stretched

dx Small displacement

This gives the elastic Potential Energy expressed in Joule.

Answer 2

Explanation:

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

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

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

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

• 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.

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