Finding two locations where power out is collision problem

[Samuelb88]

Homework Statement

A small object is shot into a block that is attached to an ideal spring. After the collision, the block moves over a horizontal surface for which it has friction. The masses of the two objects are given as well as the spring constant of the spring k, the initial velocity of the small object and the coeff. of kinetic friction between the block and the surface.

m (small) = .250 kg
M (big) = 3.75 kg
k = 180Nm^-1
v=145m/s
u_k = .600

(B) Determine two locations of the block (assume that it moves back to the left after it has reached its point of maximum spring compression) when the power output of the spring is 120. Watts.

Homework Equations

Power p = (kx)(v)

The Attempt at a Solution

This was part b of the problem. Part a was to find the max compression of the spring which is x_c (max comp) = 1.23 m.power = 120 W = kx * vx

I used the work-energy theorem to find an expression for the velocity from x_c (max comp) to the point x where the power = 120 W. I ended up with this equation

$$K_2 = k \int_{x_c}^{x} x dx - f_kx$$

$$= v = (45x^2-11.75x-68.05))^(^1^/^2^)$$

I substituted the value of v into the expression for power which gives

$$120 = 180x(45x^2-11.75x-68.05))^(^1^/^2^)$$

squaring both sides, i solve for x and get

x_1 = .59m
x_2 = 1.36
x_3 = negative value.all of which are not correct.

my professor provides one answer and that is the power output of the spring is 120 W at x = .091 m.

i don't quite understand what I'm doing wrong.

 

[willem2]

There's a problem with $f_k x$ in your equation for $K_2$ The work done by fricton should be 0 when $x = x_c$