Finding Max Compression of Spring with 500g Block

AI Thread Summary
A 500 g block slides down a frictionless track from a height of 2.00 m and compresses a spring with a stiffness constant of 20.0 N/m upon impact. The maximum velocity of the block at the bottom is calculated to be 6.26 m/s, leading to a maximum spring compression of 0.99 m. The solution confirms that equating gravitational potential energy with spring energy simplifies the calculation. The correct formula used is mgh = 1/2 kx^2, resulting in the same maximum compression. The student successfully applied this method and achieved a perfect score on their test.
AnnieF
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Homework Statement


A 500 g block is released from rest and slides down a frictionless track that begins 2.00 m above the horizontal. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring stiffness constant 20.0 N/m. Find the maximum distance the spring is compressed.


Homework Equations



mgh= 1/2mv^2
1/2KA^2=1/2mvmax^2

The Attempt at a Solution



9.8(2)=1/2v^2
v^2=6.26m/s

1/2(20) A^2= 1/2(.5)(6.26)^2

A= .99 m


I did this because I figured that if I found the velocity when the box first hit the spring, it would be considered the maximum velocity since the spring is at equilibrium at that point in time. Then, since it asked for the maximum distance it would be compressed, I just found the amplitude. Is this right? I really hope so because that is what I did on my test...Thanks for the help.
 
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AnnieF said:

Homework Statement


A 500 g block is released from rest and slides down a frictionless track that begins 2.00 m above the horizontal. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring stiffness constant 20.0 N/m. Find the maximum distance the spring is compressed.

Homework Equations



mgh= 1/2mv^2
1/2KA^2=1/2mvmax^2

The Attempt at a Solution



9.8(2)=1/2v^2
v^2=6.26m/s

1/2(20) A^2= 1/2(.5)(6.26)^2

A= .99 mI did this because I figured that if I found the velocity when the box first hit the spring, it would be considered the maximum velocity since the spring is at equilibrium at that point in time. Then, since it asked for the maximum distance it would be compressed, I just found the amplitude. Is this right? I really hope so because that is what I did on my test...Thanks for the help.
Correct. But it would have been easier simply to equate gravitational potential with spring energy.

mgh = \frac{1}{2}kx^2

x = \sqrt{2mgh/k} = \sqrt{2*.5*9.8*2/20} = .99 m.

AM
 


Thanks. I was going to do it like that but my mind was convinced that it had to be more difficult than that.
 


WOOOO HOOO! I did it right and I got 100 on my test! What an amazing day. :)
 
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