Dynamics/ Simple Harmonic motion help

[SpIIIdD]

Hello everyone, I'm new here.
I'll just post my question- hopefully someone can answer.

A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 n/m. The pulley is frictionless. The block is released from rest when the spring is un-stretched. The block moves 20 cm down th incline before coming to rest. Find the coefficient of friction between the block and incline.


m= 2 kg
d (delta x)= 20 cm (or 0.2 m)
k= 100 N/m
angle to the horizontal, theta= 37 deg.
coef of Fk= ?




a relevant equation, i think, for this problem is coef of Fk= Fk/Fn (normal force) but I'm not 100% sure about anything else.



So first i tried to draw the full body diagram of the forces of that are acting on the block.
The mg force points straight down, the kinetic frictional force should point diagonally to the left and the normal force points diagonally upward (parallel to the surface of contact where the block rests).
However, I'm not exactly sure what the force going diagonally to the right and downward- opposite to the force of kinetic friciton going opposite the direction at movement. I thought maybe that the force would be the force exerted by the string, Fx, but when i conceptually try to calculate it (with the formula Fx= -kx or kx ) i seem to obtain 0 (since the spring is being uncompressed and therefore returning to equilibrium, k should equal 0? (and multiply then by x- 0.2- and the answer is zero or have i got it wrong)?

I'm just generally confused as to how to continue with this ques and am wondering maybe there's something worng with my reasoning. I know that what I've done should be folowed up by solving a system of equations to get the final value but I'm not sure how to go about doing it.

Help?

 

[kuruman]

The mg force points straight down, the kinetic frictional force should point diagonally to the left

I hope you mean parallel to the incline.

and the normal force points diagonally upward (parallel to the surface of contact where the block rests).

You mean perpendicular to the surface of contact. That's what "normal" means.

However, I'm not exactly sure what the force going diagonally to the right and downward- opposite to the force of kinetic friciton going opposite the direction at movement.

What force would that be?

Anyway, I think it would be easier to tackle this problem by using the work-energy theorem or total (not just mechanical) energy conservation.

 

[SpIIIdD]

I hope you mean parallel to the incline.

You mean perpendicular to the surface of contact. That's what "normal" means.

What force would that be?

Anyway, I think it would be easier to tackle this problem by using the work-energy theorem or total (not just mechanical) energy conservation.

ok, so i drew two different free- body diagrams- one for the block statically in position and another one for it in movement. I derived the formula to determine thr coef of kinetic friction: mewk= Fk/FN. I'm not exactly sure how to apply the work energy theorem to solve this problem since I'm not looking for work and there isn't the coef of kinetic friction in the formula (?). Particularly I'm not sure how to utilize the spring constant and the known displacement to solve the problem.
Can someone give me a further hint?

 

[kuruman]

The work-energy theorem can be used for things other than just finding work. Don't forget that friction does work therefore μ_k appears in the work done by friction, so if you apply the work-energy theorem you will have an equation with μ_k in it.

What does the work-energy theorem say? What is an expression for it? How many forces do work and what are the corresponding expressions?

 

[rwisz]

Hello! It looks like you have a good understanding of the concepts involved in this problem, but you may just need a bit more guidance in applying them. Let's break this down step by step.

First, you are correct in identifying the relevant equation for finding the coefficient of friction. Remember that the normal force is equal to the component of the weight (mg) that is perpendicular to the incline, so you can use trigonometry to find the normal force in terms of the weight and the angle of the incline.

Next, let's consider the forces acting on the block. As you mentioned, there is the weight (mg) acting straight down, the normal force (Fn) acting perpendicular to the incline, and the kinetic friction force (Fk) acting parallel to the incline. There is also the force exerted by the spring (Fx) which you correctly identified as being opposite to the direction of motion. However, this force is not equal to 0. Remember that the spring is initially compressed, so when it is released, it will push the block in the direction of motion. This force is equal to kx, where k is the spring constant and x is the distance the spring is stretched or compressed. In this case, x is equal to the distance the block has moved down the incline, 0.2 m.

Now, we can set up our system of equations to solve for the coefficient of friction. We know that the sum of the forces in the x-direction must be equal to the mass times the acceleration in the x-direction (which is 0, since the block is not moving in that direction). So we have:

Fx - Fk = 0

We also know that the sum of the forces in the y-direction must be equal to the mass times the acceleration in the y-direction (which is also 0). So we have:

Fn - mg = 0

Now we can substitute in our expressions for Fx and Fn:

kx - Fk = 0
Fn - mg = 0

Next, we can use the equation you identified, coef of Fk = Fk/Fn, to solve for the coefficient of friction. We can rearrange the first equation to solve for Fk, and then substitute into the second equation:

Fk = kx
Fn = mg

coef of Fk = (kx)/Fn = (kx)/(mg)

Finally, we