Friction Problem

[elementcollector1]

Have been working on these for days now, and keep coming to the same answers. The problem:

 

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. (Use any variable or symbol stated above along with the following as necessary: g.)

d=(-mgμ_scos(θ)-mgsin(θ))/k

(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.

???

 

Part a is right. Part b, I have no idea where to start. I know I should set "d" equal to 0, but what do I solve for from there?

 

Other problem:

 

The potential energy of a 3.9-kg object constrained to the x axis is given by U = 2x² − x³ for x ≤ 3.0 m and U = 0 for x ≥ 3.0 m, where U is in joules and x is in meters, and the only force acting on this object is the force associated with this potential-energy function.

(a) At what positions is this object in equilibrium? (Enter your answers from smallest to largest.) x₁ = 0 m x₂ = (4/3) m

 

© Discuss the stability of the equilibrium for the values of x found in Part (a).

point x₁ is in stable equilibrium

point x₂ is in unstable equilibrium

(d) If the total mechanical energy of the particle is 11 J, what is its speed at x = 4.4 m?

 

Because potential energy is equal to 0 when x is greater than 3 (given above), and mechanical energy normally equals kinetic plus potential energy, mechanical energy in this case equals kinetic energy, given by KE=(mv²)/2. So, I set up the math as follows:

11=3.9v²/2

22=3.9v²

(22/3.9)=v²

v=sqrt(22/3.9)= 2.375 m/s

This answer is not accepted, even when negative. Why?

Edited November 14, 2013 by elementcollector1

[imatfaal]

ok - part b is weird (or I am missing something obvious - this is much more likely)

 

1st thought. Even if we discount friction the block should not be moving just before zero contraction as there is a the mgsin(theta) component of gravity acting down the slope and you need the force transmitted through the spring of F=-kx to counter that force - otherwise no movement.

 

2nd thought. You have a net force [mu_k*Normal and mgsin(theta) against -kx ]and a mass the moment the block starts to move (this force is partially dependent on distance x ). You know the block accelerates and decellerates such that velocity at x=0 v=0. This is enough to work out your one unknown mu_k but it is a beast of a calculation.

 

3rd thought. Can you use energy conservation - initially I thought not because friction is not a conservative force? You have change in U_spring, gain in U_grav,

On the second question I am equally stumped. Have you tried 4.265m/s (which would be the answer if they had forgotten the range of the PE)? Just by the by the question asks for speed not velocity so no way it can be negative - speed is a scalar and velocity is the vector. Is there a proviso for entering a specific number of DP or SF?

 

Unfortunately I think you will have to wait till one of the real physicists strolls by

[swansont]

I think I'd attack part b in terms of work and energy.

[studiot]

 

The block stops moving just as the extension of the contracting spring reaches zero.

 

 

First question

Yes swansont is correct.

 

The fact that the extension = zero means that the tension in the spring has reduced from the limiting tension (T) at the initial position to zero. This limiting tension has already been worked out in part A.

 

Since the movement was slow we assume no dissipation due to friction (!?) so all the strain energy from the spring is used in raising the block from its initial rest position up a height equal to dsin(theta) adding to its potential energy.

 

Thus PE gained by block = mg d sin(theta)

 

Strain Energy lost by spring = 0.5 T d

 

Second question

 

x ≤ 3.0 m and U = 0 for x ≥ 3.0 m,

 

Something is wrong here. You cannot have 3 as both greater than/= and less than/=.

 

In any case the formula for U give U = -9 at x=3. U corsses the x axis and become negative at x=2 so are you sure the second condition is not x>2?

Edited November 14, 2013 by studiot

[imatfaal]

But surely the energy lost through heat via friction is not related to the velocity ie W=F.d

[swansont]

Since the movement was slow we assume no dissipation due to friction (!?)

 

No, you don't want to do that. The coefficient of kinetic friction is contained in the equation for work done by friction.

 

The work will account for the difference between the changes in gravitational and spring PE.

[elementcollector1]

I found online the equation (μ_k=f/N) where N, in this case, would be the normal force (mgcos(θ)) and f would be equal but opposite to the coefficient of static friction (because the block has stopped moving, meaning the force on the block in the upwards direction equals the static friction resisting it. However, when I input μ_s/mgcos(θ), negative or positive, neither answer is accepted. What am I doing wrong here?

Edited November 14, 2013 by elementcollector1

[studiot]

Question 1 Part b Rethink.

 

I did suggest the frictional dissipation be taken as zero since I though there was insufficient information given to calculate it.

 

However on reflection I see that you can simply add another term for work done (a la swansont) against the sliding friction.

 

So strain energy of the spring = gain in potential energy of the block plus the work done against sliding friction.

 

Tis second work term is simply force (F_sliding* distance) = d = extension, and F_sliding = coefficient of sliding friction times normal reaction,.

 

When the energy equation is set up it only contains the terms required plus the coeficient of sliding friction. So this can easily put put in terms of the other terms.

 

You should be able to complete the question from here.

Edited November 15, 2013 by studiot

[elementcollector1]

Talked to my teacher about this. He said to use the following equations:

W=Fd

F=μ_kN

 

So,

W=μ_kNd

=μ_k(mgcos(θ))((-mgcos(θ)μ_s-mgsin(θ))/k)

 

From the above post, it seems like 'd' denotes work - but isn't it displacement, making work a different quantity? Since F=kd (Hooke's Law), couldn't W (being Fd) be kd²? I tried solving for the coefficient of kinetic friction as such, and found (μ_k=-μ_s- tan(θ)). This is also wrong.

Edited November 15, 2013 by elementcollector1

[studiot]

I am sorry we seem to be in widely separated time zones, but you must have a genuine interest as you have returned several times to this.

 

This may catch you quickly but I will return with a more complete post later.

 

The most important thing to understand is that the energy (or work done) in stretching a spring is not force x distance. It is One half of that.

 

E_spring = 0.5 x Force x Distance.

Edited November 15, 2013 by studiot

[swansont]

I found online the equation (μ_k=f/N) where N, in this case, would be the normal force (mgcos(θ)) and f would be equal but opposite to the coefficient of static friction (because the block has stopped moving, meaning the force on the block in the upwards direction equals the static friction resisting it. However, when I input μ_s/mgcos(θ), negative or positive, neither answer is accepted. What am I doing wrong here?

 

The block is moving during the whole extent of the problem — the solution applies at the instant it starts moving and ends the moment it stops. μ_s entersthe problem because you will use the d from part a of the problem, which is the displacement. As studiot points out, the PE for a spring is 1/2 kx²