How to get the coefficient of kinetic friction

[Fatima Hasan]

Homework Statement

Homework Equations

W = ΔUg + ΔkE + ΔUs
KE = 0.5 m v^2
Ug = m g h
Us = 0.5 k x^2

The Attempt at a Solution

k = 60 N/m vi=0 m/s (" the block is released from rest ") xi= 0 m/s ( "the spring is unstreached") vf= 0 m/s xf= 0.2 * sin 37 = 0.2*0.6 = 0.12 m d= 0.2 m hf=0 hi= 0.2*sin37 = 0.12m m= 2 kg μκ = ?
W = Fκ d cos θ
Fk = FN μk
Fk = mg cos θ ( mg must be decompose)
Fk = 2*10*0.8 = 16 N
so , W = 16 * 0.2 * cos 180 * μκ = - 3.2 μκ
W = ΔUg + ΔkE + ΔUs
since the velocity is constant , ΔkE = 0 --> W = Δ Us + ΔUg
- 3.2 μκ = 0.5 k xf^2 - 0.5 * k xi^2 + mghf - mghi
- 3.2 μκ = 0.5 * 60 * ( 0 )^2 - 0.5*60*( 0.12 )^2 + 2*10*0 - 2*10*0.12
- 3.2 μκ = -0.432 - 2.4
- 3.2 μκ = -2.832
μκ = 0.885
This is my tried , but I don't know where is my mistake.
Any help would be greatly appreciated !

 

[kuruman]

The statement of the problem mentions one block on the incline, but the figure shows an additional block on the horizontal surface. Are we to assume that this second block has the same mass as the first one? What about friction between it and the horizontal surface?

 

[Fatima Hasan]

The statement of the problem mentions one block on the incline, but the figure shows an additional block on the horizontal surface. Are we to assume that this second block has the same mass as the first one? What about friction between it and the horizontal surface?

I think it's the same object , the block was on the horizontal surface and then it falls down on an incline plane as it shown in the figure.
I calculated the work done by the frictional force . W = Fκ d cos θ , Fκ = FN μκ , since the block will move on the incline , we should decompose mg into its components . So, Fk = μκ * mg cos θ ( the angle between the frictional force and the displacement = 37° )

 

[haruspex]

- 3.2 μκ = 0.5 * 60 * ( 0 )^2 - 0.5*60*( 0.12 )^2 + 2*10*0 - 2*10*0.12

You have used the wrong value for the spring extension.

The diagram is very strange. The block is never on the horizontal surface.

 

[Fatima Hasan]

You have used the wrong value for the spring extension.

The diagram is very strange. The block is never on the horizontal surface.

So , ΔUs = 0, because the spring extension isn't change ?

 

[haruspex]

So , ΔUs = 0, because the spring extension isn't change ?

No.
The spring is initially relaxed. How far does the block slide? What is the change in the spring extension?

 

[Fatima Hasan]

No.
The spring is initially relaxed. How far does the block slide? What is the change in the spring extension?

Got it , thank you .