• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter PoohBah716
  • Start date
  • #1

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
 

Attachments

Answers and Replies

  • #2
54
10
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
 
  • Like
Reactions: BvU
  • #3
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 ????
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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.
 
  • #5
54
10
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.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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.
 
  • #7
BvU
Science Advisor
Homework Helper
2019 Award
13,019
3,009
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 ?
 

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

  • Last Post
Replies
20
Views
2K
Replies
7
Views
982
  • Last Post
Replies
1
Views
3K
Replies
9
Views
621
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
473
Replies
1
Views
2K
Top