How many trips will the box make before coming to rest?

[octahedron]

Homework Statement

A 5.0 kg box slides down a 5-m-high frictionless hill, starting from rest, across a 2-m-long horizontal surface, then hits a horizontal spring with spring constant 500 N/m. The other end of the spring is frictionless, but the 2.0-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.25.

(a) What is the speed of the box just before reaching the rough surface?
(b) What is the speed of the box just before hitting the spring?
(c) How far is the spring compressed?
(d) Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

Homework Equations

W_nc = delta-E

The Attempt at a Solution

So I actually solved (a)-(c) using non-conservative work and delta energy. My solutions for (a)-(c), if that would help, were

(a) 9.9 m/s
(b) 9.39 m/s
(c) 0.959 m

However I honestly haven't a clue as to how to approach (d) besides the "obvious" long way by going through the velocities over and over again until it reaches zero. Any elegant way here?

Thanks in advance!

 

[learningphysics]

Yes, use energy... the work done by friction is the same each time the object crosses the rough surface. Use the initial energy... and the work by friction per trip across the rough surface... to get the number of trips before energy of the box becomes zero.

 

[Hells_Kitchen]

I think its even a better idea if you use the kinetic energy of the box just before hitting the rough surface since you are including the first crossing too.
Like learningphysics said:
- Find you how much work it takes each time (its going to be the same) the box crosses the rough surface
W = f*d = mu*N(normal)*d
- Use the kinetic energy of the box just before hitting the surface (Ek = mv^2/2)
Then C*W = Ek where C is the constant (number of times it crosses the rough surface)

 

[Omega_sqrd]

I was having trouble with part d of this problem, as well. Thank you Hells_Kitchen and LearningPhysics for your help.

However, I'm still confused. Why is it that we don't need to worry about the work the spring does or the work that gravity does? Why is it that Wnet is just the work done by friction?

Thank you!

 

[octahedron]

Notice that the work done by the frictional force is not merely W_{net} but the work done by nonconservative forces W_{nc}, and since the spring and gravitational forces are not nonconservative, they are not factored in the calculation.

Then it follows that W_{nc} = \Delta E, and you use energy done throughout the work completed to obtain the \total\ distance traveled across the rough surface.

Thanks everybody; I used learningphysics's method and got the correct answer, which is 10 times.

 

[Omega_sqrd]

Thank you, Octahedron. It all makes sense now.