# Is the textbook wrong or am I?

## Homework Statement

A vertical spring (ignore its mass), whose spring constant is 875 N/m is attached to a table and is compressed down by .160 m. (a) What upward speed can it give to a .380 kg ball when released?

## Homework Equations

Conservation of Energy using 1/2 k x^2 for Uspring.

## The Attempt at a Solution

I get 7.68 m/s for the velocity and the book gets 7.47 m/s. Wanted to see who was right and if I'm doing something wrong.

LowlyPion
Homework Helper

## Homework Statement

A vertical spring (ignore its mass), whose spring constant is 875 N/m is attached to a table and is compressed down by .160 m. (a) What upward speed can it give to a .380 kg ball when released?

## Homework Equations

Conservation of Energy using 1/2 k x^2 for Uspring.

## The Attempt at a Solution

I get 7.68 m/s for the velocity and the book gets 7.47 m/s. Wanted to see who was right and if I'm doing something wrong.

"Upward" also means against gravity. You should also figure as an adjustment the m*g*h over the displacement of the acceleration.

So you're saying you got the book's answer?

I used the conservation of energy subbing values for spring and taking into account y=0 when crossing the original spring length. I'm quite sure I have the right answer and the book's is wrong but I just want to verify.

LowlyPion
Homework Helper
So you're saying you got the book's answer?

I used the conservation of energy subbing values for spring and taking into account y=0 when crossing the original spring length. I'm quite sure I have the right answer and the book's is wrong but I just want to verify.

I'm just saying that

mv2/2 = kx2/2 - m*g*x

Last edited:
You are wrong, your book is right, Pion is right: (except I would not use both "h" and "x", there is only one vertical distance in the problem)

LowlyPion
Homework Helper
You are wrong, your book is right, Pion is right: (except I would not use both "h" and "x", there is only one vertical distance in the problem)

Thanks for the catch. Of course h and x are the same.

I edited the previous post to be correct now.