Can't find mistake on Spring/block Energy problem

[falco451]

Homework Statement

A 1.93 kg block situated on a rough incline with 38.2 angle is connected to a spring of negligible mass having a spring constant of 90 N/m.
The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 10.9 cm down the incline before coming to rest.

Homework Equations

W = (fk cos 180)
x= -$$\mu$$k(mg cos theta)
Wnc = ( KE + PEg + PE s)f - (KE + PEg + PEs)i

The Attempt at a Solution

I used the equations and combined them to get:
-$$\mu$$k(18.194N cos 38.2) (.109) = 0 + (18.194)(-.109sin 38.2)
+ .5(90 N/m)(.109^2)

For my answer I got $$\mu$$k = .4438645593
It says that my answer is wrong and I can't find what I did wrong. I recalculated several times and still got the same answer. Can you help me? Thank you!

 

[falco451]

I retried calculating, and I got a different answer that was close to the one I had already gotten, but it was still wrong.

 

[ogb p]

As a scientist, it is important to double-check our calculations and make sure we are using the correct equations and values. In this case, it seems like you may have made a mistake in your calculations. First, let's review the given information and equations.

The block has a mass of 1.93 kg and is on a rough incline with a 38.2 degree angle. The spring has a spring constant of 90 N/m. The block is released from rest and moves 10.9 cm down the incline before coming to rest.

The equations we need to use are:

1. Net work done on the block = Change in kinetic energy
2. Work done by friction = Change in potential energy
3. Net work done on the block = Change in kinetic energy + Change in potential energy

Let's start by calculating the net work done on the block. We know that the change in kinetic energy is equal to zero since the block is initially at rest and comes to rest at the end. Therefore, the net work done on the block is also equal to zero.

Next, we can calculate the work done by friction. We know that the block moves 10.9 cm down the incline, which is equal to 0.109 m. The normal force on the block is equal to mg cos(theta), which is 18.194 N. Therefore, the work done by friction is equal to mu_k * (18.194 N) * (0.109 m) * cos(180 degrees) = -1.984 mu_k.

Lastly, we can calculate the change in potential energy. We have two types of potential energy in this problem - gravitational potential energy and spring potential energy. At the beginning, the block is at a height of 0.109 m above the ground, so the initial gravitational potential energy is equal to mgh = (1.93 kg) * (9.8 m/s^2) * (0.109 m) = 2.012 J. At the end, the block is at a height of 0.109 m above the ground, so the final gravitational potential energy is also equal to 2.012 J.

For the spring potential energy, we can use the equation PE_s = 0.5 * k * x^2, where k is the spring constant and x is the displacement of the block. In this case, the displacement