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Work of spring

  • Thread starter mybrohshi5
  • Start date
  • #1
365
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Homework Statement



A 3.00 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 468 N/m) the other end of which is fixed. The ladle has a kinetic energy of 10.00 J as it passes through its equilibrium position (the point at which the spring force is zero).

At what rate is the spring doing work on the ladle when the spring is compressed by 10.0 cm and the ladle is moving away from the equilibrium position?

Homework Equations



K = 1/2mv2

W = 1/2kx2

F = -kx

P = F*v

The Attempt at a Solution



I found the velocity

10 = 1/2(3)(v2)
v = 2.58

I found the force next

F = -468(.1)
F = -46.8

Then i found the power which is the rate of the spring doing work on the ladle

P = -46.8(2.58)
P = -120.74 W

Can anyone tell me where i went wrong here?

Thank you
 

Answers and Replies

  • #2
96
0
I think you're not supposed to use the speed for this calculation. The speed you calculated is for the equilibrium state, but thats not the speed of the ladle when the spring is compressed by 10cm.
 
  • #3
365
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If i dont use speed do you have any suggestions on how to find the answer cause i am drawing a blank right now...
 
  • #4
123
1
Pinsky's right. You are on the right track and you just have to find its speed when it's compressed by 10cm using the kinetic energy formula.
 
  • #5
365
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Does this maybe seem right to find my velocity i need.

W = 1/2 k x^2

W = 1/2(468)(.1^2)

W = 4.68 J

W = K = 1/2mv^2 - 1/2mv_i^2

4.68 = 1/2(3)(v^2) - 0

v = 1.767 m/s
 
  • #6
123
1
Again, it's the right idea. You made two mistakes. Work done should be negative. Think about why that is. And your v_i should be 2.58 where you got from 10J of energy.
 
  • #7
365
0
Thanks i got the final velocity now :)

-4.68 = 1/2(3)(v) - 1/2(3)(2.58)

v = 1.88 m/s

then from here i am not sure what force to multiply this velocity by.

is it F = -kx

F = -468(.1)

F = -46.8 J

so

P = -46.8(1.88) = -88.01 W
 
  • #8
123
1
I think you've got the right answer. Just remember, the equation F= -kx, where x is the distance relative the equilibrium position.
 
  • #9
365
0
Thank you for the help. I just tried -88.01 but my online homework said it was wrong.. I dont see anything wrong hmmm.. :(
 
  • #10
123
1
You are welcome. But I don't anything wrong either. Maybe it's the rounding?
 
  • #11
123
1
Does this maybe seem right to find my velocity i need.

W = 1/2 k x^2

W = 1/2(468)(.1^2)

W = 4.68 J

W = K = 1/2mv^2 - 1/2mv_i^2

4.68 = 1/2(3)(v^2) - 0

v = 1.767 m/s
I found your mistake. Work should be -2.34 J.
 
  • #12
365
0
ahhh! thank you so much. i really appreciate it :)
 

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