Block on a spring on a horizontal surface with friction?

  • #1
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Homework Statement


A 3.0 kg block is held against a spring with a force constant of 125 N/m. The spring is compressed by 12 cm. The ice is released across a horizontal plank with a coefficient of friction of 0.1

A) Calculate the velocity of the block just as it leaves the spring. Assume the friction between the plank and block is negligible until the moment the block leaves the spring.

B) Determine the distance the block travels after it leaves the spring.

Homework Equations


Ek = 0.5mv^2
Ee = 0.5KX^2
W = Fdcostheta


The Attempt at a Solution



I did it just how I would do a problem except I threw in the work formula in there so:

Ek1 = Ee2 + (Fdcos180)

After cancelling the 0.5 in each equation: mv^2 = kx^2 + (fdcos180)

3v^2 = (125 x 0.12^2) - (0.1 x 3 x 9.81 x 0.12)

3v^2 = 1.45

V = 1.44 m/s

This is wrong because the answer key says the answer is 0.77 m/s
 

Answers and Replies

  • #2
PhanthomJay
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The question has 2 parts and you are mixing them together I think and approaching it wrong. The first part assumes no friction in between the initial release of the the spring from rest and the point where the block leaves (loses contact with) the spring. Try again using energy conservation. What is the potential energy of the spring when the block loses contact with it?
 
  • #3
haruspex
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After cancelling the 0.5 in each equation: mv^2 = kx^2 + (fdcos180)
There was not a factor of 0.5 in the fdcos180 term. But as PJ points out, you don't need it.
3v^2 = 1.45

V = 1.44 m/s
Try that step again.
 
  • #4
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There was not a factor of 0.5 in the fdcos180 term. But as PJ points out, you don't need it.

Try that step again.
Woops, I messed up solving for v but I end up with 0.7 after I solved correctly? Where else could I have messed up that accounts for the .07 difference?
 
  • #5
PhanthomJay
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Woops, I messed up solving for v but I end up with 0.7 after I solved correctly? Where else could I have messed up that accounts for the .07 difference?
Please show how you arrived at that incorrect result.
 
  • #6
haruspex
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Woops, I messed up solving for v but I end up with 0.7 after I solved correctly? Where else could I have messed up that accounts for the .07 difference?
You must be still subtracting for energy lost to friction. As PJ pointed out, the question says:
Assume the friction between the plank and block is negligible until the moment the block leaves the spring.
If that does not explain it, please post all your working now.
 
  • #7
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The question has 2 parts and you are mixing them together I think and approaching it wrong. The first part assumes no friction in between the initial release of the the spring from rest and the point where the block leaves (loses contact with) the spring. Try again using energy conservation. What is the potential energy of the spring when the block loses contact with it?
Ohh okay, I got 0.77 when taking out friction but now how would I solve for the distance in part b? Is this where I plug in friction?
 
  • #8
PhanthomJay
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Ohh okay, I got 0.77 when taking out friction but now how would I solve for the distance in part b? Is this where I plug in friction?
yes give it a go.
 
  • #9
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yes give it a go.
Where exactly would I solve for the distance? Isn't the distance in (Fdcostheta) just equal to 0.12?
 
  • #10
haruspex
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Where exactly would I solve for the distance? Isn't the distance in (Fdcostheta) just equal to 0.12?
No, the 0.12 was the distance the block moved up to the point where it left the spring. You now want to find out how much further it goes.
 

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