# Homework Help: Spring and wood block problem

1. Feb 25, 2005

### whereisccguys

A 171 g wood block is firmly attached to a very light horizontal spring, as shown in the figure below.
The block can slide along a table where the coefficient of friction is 0.306. A force of 20.9 N compresses the spring 17.2 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

i know the equation is suppose to be 1/2k(x^2-D^2)=(Uk)mg(x+D)
where D is the distance the block will travel....
and then set the equation equals to D = x - (2Ukmg)/D
i plug in all the numbers and i still get the wrong answer...
i think my K constant is wrong i thought it was just k = 20.9 N/.172 m = 121.51

btw my wrong answer is 8.7596 m

can someone help me?

#### Attached Files:

• ###### dgian0834.gif
File size:
3.7 KB
Views:
150
Last edited: Feb 25, 2005
2. Feb 25, 2005

### Gamma

Frictional force is Umg. Not Ukmg.

Also Check your calculations. There can not be a D in the denominator.

I am getting,

D = x - (2Umg)/k

3. Feb 25, 2005

### MathStudent

I think he meams that Uk is the kinetic coeffiecient of friction (where the k is meant to be a subscript, not the spring constant).
but other than that, I got the same equation as gamma.

like gamma said, check your calculations, and also check your units: g -> kg, etc...

4. Feb 25, 2005

### whereisccguys

yea mathstudent is right i was talkin about kinetic coefficient and yea i accidently typed the wrong thing for the equation... mathstudent was right i forgot about the units... i used g instead of kg... thanks a lot guys!

Last edited: Feb 25, 2005