# I just took a test and I want to know if I did this problem right? Springs+ Vibration

## 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.

## Answers and Replies

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Andrew Mason
Science Advisor
Homework Helper

## 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.
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. :)