Box dropped on spring

  • Thread starter jensson
  • Start date
  • #1
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Homework Statement


A 2.40 kg block is dropped onto a spring and platform of negligible mass. The block is released a distance of 5.00 m above the platform. When the block is momentarily at rest, the spring is compressed by 25.0 cm. Find the speed of the block when the compression of the spring is only 15.0 cm.

I need some help setting this up...


Homework Equations



F= -kx

PE = mgh

KE = 1/2mv^2



The Attempt at a Solution



I started off with mgh = -kx + 1/2mv^2 but I don't know what the value of k is so I can't solve for v (I don't have the answer, but I don't think the letter k would be part of it). I know the distance between where the block starts and compression at 15.0cm is 5.15cm so I think I use that for x? really I just have no idea how to do this and I can't find an example in my book of a block falling on a spring. help!
 

Answers and Replies

  • #2
You must also consider the energy stored in the spring
[tex]
E = \frac{1}{2}kx^2
[/tex]
The total potential energy lost by the box during the fall, is equivalent to the energy stored in the spring at the moment when the box is at rest. This will allow you to calculate k.

Then use the total energy as
[tex]
E_T=mgh+\frac{1}{2}mv^2+\frac{1}{2}kx^2
[/tex]
to find the speed of the block when x = 15cm
 
  • #3
I started off with mgh = -kx + 1/2mv^2 but I don't know what the value of k is so I can't solve for v (I don't have the answer, but I don't think the letter k would be part of it). I know the distance between where the block starts and compression at 15.0cm is 5.15cm so I think I use that for x? really I just have no idea how to do this and I can't find an example in my book of a block falling on a spring. help!

Firstly in you equation mgh = -kx + 1/2mv^2 you are equating force with energy. It should be 1/2 kx2.

Also, you can calculate k from the given data. Just try it!
 

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