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

[PoohBah716]

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?

Homework Equations

W=(1/2)KA^2

The Attempt at a Solution

426 n/m and 3.7 kg

 

[AbhinavJ]

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

 

[PoohBah716]

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]

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.

 

[AbhinavJ]

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]

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]

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 ?