How Does a Spring's Work Relate to Frictional Forces in Motion?

[gagga5]

A relaxed spring with spring constant k = 60 N/m is stretched a distance di = 59 cm and held there. A block of mass M = 7 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/9.

Ok, so there is three questions.

a) In moving from the initial to the final position, by how much has the kinetic energy of the block changed?

- I found out this is zero, since v = 0, for initial and final.

(b) What is the work done by the spring?

- This question is driving me insane. I know that potential energy equation for the spring is U = 1/2 k(Sf-Si)^2.
So I tried with 1/2*60*(59)^2 - 1/2*60*(59-59/9)^2 = 103141
it was not the answer. also i tried with negative sign, I tried adding them, tried all kinds of possibilities, but all of them were not the answer. I think I should use different way to solve this problem.

(c) What is the magnitude of the total work done by the frictional force?
I guess I can figure this out once i know about the work done by the spring...

(d) What is the magnitude of the frictional force on the block?
This one could probably be solved once I know about c..

So my main question is how do I figure out the work done by the spring.

 

[Doc Al]

(b) What is the work done by the spring?

- This question is driving me insane. I know that potential energy equation for the spring is U = 1/2 k(Sf-Si)^2.

At any point, the amount of PE stored in the spring is given by U = 1/2 k(x)^2, where x is the amount that the spring is stretched from its relaxed position. In this case, the change in spring PE would be 1/2 k (Sf^2 - Si^2).