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

• PoohBah716
In summary, Abhi invited OP to find out for himself, but then spoiled the fun by providing the answer.
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?

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
AbhinavJ said:
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 ?

PoohBah716 said:
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.

AbhinavJ said:
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.

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

1. What does the equation W=(1/2)KA^2 represent?

The equation W=(1/2)KA^2 represents the work done by a spring system, where W is the work, K is the spring constant, and A is the displacement of the spring from its equilibrium position.

2. How is this equation derived?

This equation is derived from Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. By integrating the force equation, we can obtain the equation for work done by a spring system.

3. What is the significance of the 1/2 factor in the equation?

The 1/2 factor in the equation represents the average force exerted by the spring over its displacement. This is because the force exerted by a spring is not constant and varies as the spring is stretched or compressed.

4. Can you explain the units of each variable in the equation?

The units of W, the work, is joules (J) which is a measure of energy. The units of K, the spring constant, is newtons per meter (N/m) which is a measure of stiffness. The units of A, the displacement, is meters (m) which is a measure of length.

5. How does this equation relate to real-life situations?

This equation is commonly used in physics and engineering to calculate the work done by a spring system in various applications such as in springs, shock absorbers, and elastic materials. It is also used in the study of simple harmonic motion and oscillations.

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