# Homework Help: Finding Velocity of a System with Work and Energy

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1. Nov 2, 2016

### xTheLuckySe7en

1. The problem statement, all variables and given/known data
The coefficient of friction between the block of mass m1 = 3.00 kg and the surface shown is µk = 0.400. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.50 m?

Note that this is just a generic picture of how the problem is set up. M would be m1 and m would be m2 (the ball).

2. Relevant equations

W = F ⋅ Δr = F * r * cosΘ
Fk = µk * Fn, where Fk is the force due to kinetic friction and Fn is the normal force.
∑W = ΔK, where W is work and K is kinetic energy
Ki + Ui + W - Fkd = Kf + Uf, where K is kinetic energy, U is potential energy and Fkd is the work done by friction (non-isolated system)
K = (1/2)mv^2
U = mgh (or mgΔy)

3. The attempt at a solution
So, I would like to mention that I do have access to the solution to the problem, and I believe I understand the general idea of how it works (treat both masses as a system, and then use the non-isolated system equation listed above with work set equal to zero and so on). Although, before looking at the solution I tried it myself (which turned out to be incorrect).
So, I started by stating that ∑W = Wt + Wk (Wt = work done by tension force and Wk = work done by friction), since the normal force and gravity are not doing any work here (m1 isn't moving vertically, so Δr = 0 in the y-direction).
I got Wt = (5.00kg)(9.80m/s^2)(1.50m)cos0 = 73.5J.
I found friction to be Fk = μk * Fn = (.400)(5.00kg)(9.80m/s^2) = 19.6N, since Fn = mg.
I then found Wk = (19.6N)(1.50m)cos180 = -19.6N
I found ΣW = Wt + Wk = 73.5J - 19.6J = 44.1J
I then used ΣW = ΔK. I know Ki is 0 since it states the system is initially at rest, so ΔK = Kf.
ΔK = (1/2)m1vf^2
Setting them equal to each other...
vf = √[(2*ΔK)/m1] = √[(2*44.1J)/(3.00kg)] = 5.42m/s

The actual answer is vf = 3.74m/s. Why does my attempt not work out to be the correct answer? I know the velocities of both masses are equal due to the tensile force between them and the frictionless (assumed) pulley. I tried thinking of it a different way to try to tell myself that it wasn't correct. I thought of a parallel force pulling m1 with the same force as m2g (49N), and then calculated the total work done and then relating it to the work-kinetic energy theorem to get my answer and it made sense in my head. I'm not understanding how the fact that it's a non-isolated system should get a different velocity than my answer.

2. Nov 2, 2016

### TomHart

Because the blocks are accelerating, the tension is not equal to the weight of m2. If the tension was equal to the weight of m2, m2 would be in equilibrium and not accelerating at all.

3. Nov 2, 2016

### xTheLuckySe7en

Oh, okay. Thank you for that. I would then assume that we wouldn't really be able to find out the tensile force with Newton's Second Law because you would need something for the acceleration of one of the masses, correct?

4. Nov 2, 2016

### TomHart

Check that equation.

Also, in calculating friction force, it looks like you used the wrong mass.

I generally lean towards solving these problems using free body diagrams, where you end up with 2 simultaneous equations. But if I look at this problem in terms of energy (which probably makes more sense), here is my understanding.

After the blocks have moved 1.5 meters, block 2 (5 kg) has a decreased potential energy, which can be calculated. In addition to that, both blocks have increased kinetic energy based on their final velocity (whatever it turns out to be). In addition to that, not all of the potential energy of block 2 was converted to kinetic energy; some of it was lost in the heat that was generated by the friction between block 1 and the surface. That's really all there is that I can see if you are looking at this as a system. The tension in the rope should not come into play. So if I had to write an equation for the system, I would write:

Epi + Eki + W = Epf + Ekf

Epi - Initial potential energy
Eki - Initial kinetic energy
W - Work added to system (in our problem W would be negative because it was energy leaving the system)
Epf - Final potential energy
Ekf - Final kinetic energy