# W=(1/2)KA^2 Can someone explain this equation in detail & Q

## Homework Statement

A mass resting on a horizontal frictionless surface is attached to one end of a spring. 3.6 J to compress spring 0.13 m. released at max acceleration 15 m/s. What is the spring constant and mass

I would like to dicuss this question how the equations are derived. can someone give me there explainantion of the problem attached?

W=(1/2)KA^2

## The Attempt at a Solution

426 n/m and 3.7 kg

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Try to derive it yourself. Ill give you a hint. Restoring force applied by a spring is directly proportional to its contraction/expansion in the direction to attain natural length. F=-Kx, also work done is F. dx. Try getting an answer

• BvU
Try to derive it yourself. Ill give you a hint. Restoring force applied by a spring is directly proportional to its contraction/expansion in the direction to attain natural length. F=-Kx, also work done is F. dx. Try getting an answer
I know using Hookes law can be used after finding the mass. I want to know more how W=1/2KA^2 is derived in this solution. My understanding is W is the potential energy the and A^2 is the amplitude and the reason it is square because ????

Ray Vickson
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I know using Hookes law can be used after finding the mass. I want to know more how W=1/2KA^2 is derived in this solution. My understanding is W is the potential energy the and A^2 is the amplitude and the reason it is square because ????
You have an expression for the force $F$ exerted by the spring when it is compressed to length $x$ (Hooke's Law). How would you compute the work needed to compress the spring from its natural length to a smaller length?

Note: you should make a serious attempt to understand what is being asked here, and how to do it.

Well, whatever force we apply on the spring to contract it is stored in the spring as its potential energy. So the work done in compressing the spring is F.x. Where x is the compression in the spring. As the force varies with the compression(kx) here we calculate the work done for elemenal compression dx. Work done dw=potential of spring =F. dx=-Kx.dx.

As the force is opposite to the elemental compression dW=Kxdx.
Integrating with limits from 0 to x. We get W=integration(kxdx) =1/2Kx^2.

Ray Vickson
Homework Helper
Dearly Missed
Well, whatever force we apply on the spring to contract it is stored in the spring as its potential energy. So the work done in compressing the spring is F.x. Where x is the compression in the spring. As the force varies with the compression(kx) here we calculate the work done for elemenal compression dx. Work done dw=potential of spring =F. dx=-Kx.dx.

As the force is opposite to the elemental compression dW=Kxdx.
Integrating with limits from 0 to x. We get W=integration(kxdx) =1/2Kx^2.
Good.

BvU
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2019 Award
Ray,

I have the impresssion Abhi made a good start inviting OP to find out for himself and then spoilt the fun by providing the answer !

PB: is it clear to you now ?

• PoohBah716