Energy Conservation Spring Pulling Mass Up Incline Plane

[Creepypunguy]

Homework Statement

http://imgur.com/xIckJqW
A block of mass m rests on a plane inclined at an angle θ with the horizontal. A spring with force constant k is attached to the block. The coefficient of static friction between the block and plane is μ_s. The spring is pulled upward along the plane very slowly.

(a) What is the extension of the spring the instant the block begins to move?

(b) The block stops moving just as the extension of the contracting spring reaches zero. Express μ_k (the coefficient of kinetic friction) in terms of μ_s and θ.

Homework Equations

F = -kx

E_sys = E_mech + E_therm + E_chem + E_other

G.P.E.: U = U₀ + mgh

Spring Potential Energy: U = (1/2)kx²

Kinetic Energy: K = (1/2)mv²

F_n = mgcosθ

The Attempt at a Solution

Solution For Part (a):

X forces:
(-kx) , mgsinθ , F_nμ_s
Y forces:
F_n, F_gcosθ

ƩF_x = -F_nμ_s - mgsinθ + (-kx) = 0
Solve for x:
x = (-mg/k)(μ_scosθ+sinθ)

Attempt at a solution for part (b):

E₁ = (1/2)mv₀² + (1/2)kx²
E₂ = mgh + f_kx
f_kx = F_nμ_kx = mgμ_kxcosθ

I'm just not really sure how to go about this question at all. I'm unsure about assuming that there is an initial velocity, but I'm basing that on the previous problem.

 

[Doc Al]

Solution For Part (a):

X forces:
(-kx) , mgsinθ , F_nμ_s
Y forces:
F_n, F_gcosθ

ƩF_x = -F_nμ_s - mgsinθ + (-kx) = 0
Solve for x:
x = (-mg/k)(μ_scosθ+sinθ)

Looks good.

Attempt at a solution for part (b):

E₁ = (1/2)mv₀² + (1/2)kx²
E₂ = mgh + f_kx
f_kx = F_nμ_kx = mgμ_kxcosθ

I'm just not really sure how to go about this question at all. I'm unsure about assuming that there is an initial velocity, but I'm basing that on the previous problem.

The mass is initially at rest, until the spring force is great enough to overcome static friction. So you are starting with a mass at the end of a stretched spring. What's the energy of the system at that moment?

 

[gneill]

For part (b) you can take the starting condition being the instant just before the block begins to move (so the spring is at maximum extension), and ends when the block has just come to rest. So the block is at rest at both ends, making life simpler

EDIT: Doc Al got there before me!

 

[Creepypunguy]

So the Energy at the initial point would just be the potential energy of the spring?

 

[Creepypunguy]

Ok so starting now from
(1/2)kx² = mgx(sinθ+μ_kcosθ)
Sub in x from previous solution;
(-1/2)mg(μ_scosθ+sinθ) = mg(sinθ+μ_kcosθ)
With a small amount of arithmetic becomes;
μ_k= (-1/2)(μ_s+3tanθ)

Does that final solution check out?

 

[Doc Al]

Ok so starting now from
(1/2)kx² = mgx(sinθ+μ_kcosθ)

Good.

Sub in x from previous solution;
(-1/2)mg(μ_scosθ+sinθ) = mg(sinθ+μ_kcosθ)

Get rid of that minus sign. (I should have mentioned that earlier--you had the sign of the spring force wrong.)

With a small amount of arithmetic becomes;
μ_k= (-1/2)(μ_s+3tanθ)

Redo it, after the above correction.

 

[Creepypunguy]

After the above corrections I came to a final answer of:

μ_k = (1/2)(μ_s - tanθ)

 

[Doc Al]

After the above corrections I came to a final answer of:

μ_k = (1/2)(μ_s - tanθ)

Looks good to me.

 

[Creepypunguy]

Thanks guys!

 

[gamma-guy]

E₁ = (1/2)mv₀² + (1/2)kx²

How did you get that equation?

 

[squirlyskiing]

How did you get that equation?

This is from E1 = Ke + Pe (potential energy for a spring)

 

[haruspex]

This is from E1 = Ke + Pe (potential energy for a spring)

Your reply is nearly six years too late.