# Homework Help: Sprin compression

1. Aug 8, 2007

### andrewn

1. The problem statement, all variables and given/known data
Could somebody help walk me through this problem?:
A spring has its right end fixed and is installed on a horizontal table so that the free end, in equilibrium, is at x= 3m. A 1.65 kg block coming from the left slides along the table. When it passes the origin, it is moving at 5.58 m/s. It strikes the spring, compresses it momentarily, and is then sent back toward the left, where it eventually comes to rest at the point x= 1.5m. The coefficient of kinetic friction betweeen the block and the table is .300. By what distance was the spring compressed?

2. Relevant equations
F(of kinetiic friction)=(coefficeint of kinetic friction)(m)(g)
KE=.5(mass)(velocity squared)

3. The attempt at a solution
KE=.5(1.65kg)(5.58 m/s squared)
=25.57Joules
F(of kinetic friction)=.3(1.65)(9.8)
=4.851 N

2. Aug 8, 2007

### mgb_phys

You seem to have ended up with a force not a distance!

You had almost the correct approach, energy is conserved
First it is the KE of the incoming block.
Then some is lost in force*distance of the friction on the table.
Then it is the force*distance of the compressed spring (when the block is at rest) then it is the force*distance the retreating block travels.

3. Aug 8, 2007

### learningphysics

I'd use this idea... Work done by friction = final energy - initial energy

The distance travelled into the spring is a variable in the left side of the equation.