Calculating Spring Stretch in a Horizontal Setup

[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?

 

[Gamma]

1/2k(x^2-D^2)=(Uk)mg(x+D)

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

 

[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...

 

[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!