Conservation of energy of spring

[etothey]

Homework Statement

A 194 block is launched by compressing a spring of constant 200 N/m distance of 15cm . The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction 0.27 . This frictional surface extends 85cm , followed by a frictionless curved rise.
No friction on the curved rise, only on horizontal plane.
So basically, box has potential energy in spring, moves over friction surface, then up the hill with no friction, comes back against friction surface, hits the spring, goes to friction surface, and up the hill again, and so on.

Homework Equations

kx^2/2=mv^2/2
Ffriction=N*u=mg*u
Vf^2=Vo^2-2ax

The Attempt at a Solution

kx^2/2=mv^2/2
Solve for initial velocity which is 4,8162m/s.
Solve for deaccelration mg=mau
a=-2.646m/s^2.
Then just calculating the velocity each time he exits the friction surface, moves up, comes down with same velocity, loses velocity at friction surface, hits spring, bounces back with same velocity, enters surface, loses velocity etc.
His final velocity is 0.84m/s moving towards the left at the beginning of the frictionless surface before he comes to a stop.
He will make it 16cm from the right side, in a total of 69cm from the left side.
Is this correct?
Thank you!

 

[Delphi51]

I'm having difficulty understanding the question. Is the spring really compressed a distance of 15 meters? You don't give any units, so the question is not clear. Is the block 194 kg or 194 grams?

The diagram is not shown. It matters what the shape of the hill is because the energy lost to friction depends on the distance along the slope of the hill and on its angle.

It isn't possible to find the answer with the information you have given. Can you give us the question in full - every word - and include a sketch of the diagram or at least a description?

 

[etothey]

Edited the question.

 

[Delphi51]

I used an energy approach, basically saying that the initial spring energy is converted to work done against friction:
½ k⋅x² = μmgd
The d works out to about 5 times the 0.85 m of the surface with friction. I get a little less than your 16 cm of extra distance after the 5 passes through the 85 cm. You might use the energy approach as a check on your work.

What are we trying to find?

 

[etothey]

I used an energy approach, basically saying that the initial spring energy is converted to work done against friction:
½ k⋅x² = μmgd
The d works out to about 5 times the 0.85 m of the surface with friction. I get a little less than your 16 cm of extra distance after the 5 passes through the 85 cm. You might use the energy approach as a check on your work.

What are we trying to find?

Nice, After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Solving it you're way, i get 71.7cm, this correct?

 

[Delphi51]

Looks good! I got 72 cm; only did it to 2 significant digits.