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Vertical spring physics

  • Thread starter fiziksfun
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  • #1
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An 5 kg block is fastened to the top of a vertical spring (perpendicular to the floor) with a spring constant of 1000 N/m. A 3 kg block sits on top of the 5 kg block.

The springs are pushed down so that they oscillate.

I need help finding the magnitude of the maximum acceleration the blocks can obtain while still remaining in contact. I have no idea where to begin.

Any suggestions!? Help please!
 

Answers and Replies

  • #2
Hootenanny
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HINT: Consider the forces acting on the 3kg block, which force will be zero when the blocks lose contact?
 
  • #3
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Ok so when the blocks lose contact, the force of the spring-mass will be equal to the force of gravity on the 3 kg block.

M(3)*a=-2mg, is this correct?
 
  • #4
Hootenanny
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You've still not answered my first question. What are the forces acting on the top block?
 
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The forces acting on the block are gravity and the force of the spring, correct? Or friction??
 
  • #6
Hootenanny
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The forces acting on the block are gravity and the force of the spring, correct?
Correct, but what I was trying to get at is that the force of the spring acts through the normal force exerted on the block. Hence, the net force acting on the block is [itex]N - mg[/itex]. Can you now use this information to write an equation using Newton's second law?
 
  • #7
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ma = N - mg

but what is the magnitude of the normal force !?!?!? kx !?
 
  • #8
Hootenanny
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ma = N - mg
Correct! And what do you know about the normal force when the block leaves the surface of the 5kg block?
 

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