# Conservation of Energy: Pulley and Springs

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1. Jul 23, 2017

### OlicityFangirl

1. The problem statement, all variables and given/known data
1. Two blocks are connected with a massless, non-stretchable rope, and connected to a spring that is fixed to a wall. One of the blocks is hanging from the side of the table, while the other lies on the table. The mass of each block is m=1.0 kg, spring stiffness is k=50 N/m, and the friction coefficient between the block and the table is u=0.3. The spring is originally unstretched while the blocks are held at rest. Find maximum velocity of the blocks once the system is released and the hanging block starts to go down.

*Note: I thought the description of the set up was a bit confusing. I've gotten clarification from my teacher and here it is: There is a wall with a spring attached to it. The other end of the spring is attached to "Block A," which rests on a level table. The other end of Block A is attached to the massless, non-stretchable rope, which is on a pulley. On the other end of the rope is "Block B" which is hanging.*
2. Relevant equations
Force of a Spring: F=-kx
Newton's Second Law (maybe?): F=ma
Work (maybe?): W= Fs
Work done by a spring (maybe?): W=0.5kx2

3. The attempt at a solution
I started by drawing a free-body diagram of the problem. The forces on Block A (the one on the table connected to the spring are):

-tension from the rope pulling it in the positive x-direction
-force from the spring pulling it to the negative x-direction
-friction, which opposes the direction of motion. This is presumably going to be acting the negative x-direction
-Fa=t-kx-ukmg

The forces on Block B (hanging) are:

-tension from the rope pulling it up
-gravity, pulling it down
-Fb=mg-t

Note: I'm about to briefly explain an approach that I'm 99% sure doesn't work. Please feel free to skip this paragraph. If someone could explain exactly in terms of physics why this doesn't work, I'd appreciate it!

Since the rope is massless and non-stretchable, the tensions at both ends would be the same right? I had originally attempted to solve this problem by solving for tension using Newton's Second Law and Fb=mg-t. Then I plugged in my value for tension into the Newton's Second Law equation with Fa=t-kx-ukmg. Since the accelerations of each block is the same due to the rope, I could isolate and solve for acceleration as a function of position (x). Using this method, I found that acceleration is 0 at x=0.098 m. But I'm not sure if this is where velocity is maximized since I don't know the velocity equation. Also, this approach is very Newton's Second Law heavy, which isn't what the class is focusing on right now.

Here is my current approach to the problem, a more work and conservation of energy method:

Relevant Equations for Attempt #2:

-Ugrav=mgh
-K=0.5mv2
-Uelastic=0.5kx2
-Wfriction=ukmg(s)
-Conservation of Energy: K1 + Ugrav,1 + Uelastic, 1+Wfriction=K2 + Ugrav, 2 + Uelastic, 2

Attempt #2:

Initially, the spring is unstretched, and. nothing is moving, so K1 and Uelastic, 1 are both 0. The only energy is the gravitation potential energy of Block B. After everything is released, gravitational potential energy is converted to kinetic energy and work is done on the spring, which also uses up gravitation potential energy. Some mechanical energy is also lost through friction:

Since there are two blocks of mass, m, kinetic energy of the system would be mv2. Since work done by friction is always negative, it's term is negative.

mgh1 + -mukg(s) = mv2 + mgh2 + 0.5k(s)2

I then went to isolate the kinetic energy term:

mg(h1-h2) - 0.5k(s)2 - mguk(s) = mv2

The change in height of Block B (h1 - h2) has to be equal to the displacement, s, I believe, because the blocks are connected by a pulley. So that makes the equation:

mg(s) - 0.5k(s)2 - mguk(s) = mv2

Solving for v:

[mgs - 0.5ks2 - mguks] / m = v2

{[mgs - 0.5ks2 - mguks] / m}0.5 = v

It appears the v2 is a parabolic function of displacement. I went and "looked ahead" by graphing (I'll have to show the algebra/calculus in my work) this function with the given values. I got a very small maximum velocity (0.686 m/s) at the displacement of 0.137 m. Is my work correct?

Last edited: Jul 23, 2017
2. Jul 23, 2017

### TSny

Welcome to PF!

Your work looks very good to me. I also got the same answer.

Note that as long as the net force acting on the system is nonzero and in the direction of the initial motion, the system will be speeding up. Therefore, the maximum speed occurs at the position where the net force is zero. So, as a check, you can see if the net force is zero at s = 0.137 m.