How Does Friction Affect Spring Compression and Energy Loss?

[becksftw]

Homework Statement

A horizontal uncompressed spring of constant k= 3 N/m and negligible mass lies on a frictionless floor. The right end of the spring rests against a vertical wall. A block of mass m=0.180 kg with initial velocity v0= 1.0 m/s is incident to the spring from the left. The spring in compressed and the block comes momentarily to rest.

a. Calculate the maximum compression x of the spring. (Did this, x= 0.245m)
b. Determine the velocity v of the block when the spring compression x'= 0.150m (Did this, v= 0.790 m/s)

*****Here is where I'm stuck

The initially frictionless floor is now replaced by a floor that has a coeff. of kinetic friction mu= 0.11

c. Calculate the new maximum compression x'' of the spring when the block comes momentarily to rest.

d. Determine the amount E_th of thermal energy produced in part c by frictional force.

I think I know how to do part d once I get c, but I've been stuck on part c for some time now.

Homework Equations

Fd= (Delta E_mec) + f_kd

So, I know f_k= 0.194
and deltaE= .09 ?

I'm not sure how to approach the problem, and its very important that I understand how to for my exam coming up friday. Any help is appreciated, thanks!

 

[noleguy33]

Set up your work-energy theorem so that everything is in your system.

So:

0 = deltaKtrans + deltaThermal +deltaUspring

You only have one unknown, the 'x'(or distance) value in your deltaThermal and deltaSpring expression. Solve for that x and plug in your knowns.

Once you solve for x, plug it back into your deltaThermal expression and evaluate it.

 

[becksftw]

Ok, so I get:

1.5x^2 + 0.194x + .09 =0

I end up have to take the sq root of a negative though, so I must be doing something wrong. -

sqrt(0.194^2 -4(1.5)(.09)) = -.502

 

[noleguy33]

remember the delta's are (final - initial) and the mass comes to a complete stop(i.e. it's final kinetic energy is zero)

 

[becksftw]

Oh yeah! I got 0.188m and 0.0364J which I believe are the correct answers. Thank you!