Non-Conservative Work systems coupled with friction.

[drewmacq]

1. The problem statement:

The coefficient of friction between the 3.00 kg block and surface in the figure below is 0.400. The system starts from rest. What is the speed of the 5.00 kg ball when it has fallen 1.10 m? The picture "below" depicts the 3.0 kg block on the table, attached to a string that is connected to a pulley at the end of the table and is also connected to the hanging 5.0 kg ball.

Homework Equations

PEg: mgh
KE: .5mv^2
Wnc = Change in PEg + Change in KE
Wnc (Distance that non-conservative factor is acting upon system) X (Force that this non-conservative factor is applying)

The Attempt at a Solution

So I assumed that this is a Work Non-Conservative problem because friction is involved. The problem states that the 5.00 kg ball has fallen 1.1 m, and since it is attached by a string to the 3.00 kg block, I assumed that the 3.00 kg block slid along the table 1.1 m. This would be the distance of the non-conservative factor. The force of the non-conservative factor would be equal to the Coefficient of friction X the normal force (in this case, (0.4)(3.0 kg)(9.81m/s^2). This gave me 12.9492.
Next, I assumed that the final potential energy of the 5.0 kg ball would be zero (because i defined the point 1.1 m below it's starting point as PE = 0), therefore the change in PE (PEf-PEi) would be -mgh. Since the ball is starting from rest, there is no initial KE, so I set my equation up as 12.9492 = -mgh + .5mv^2

Obviously, I come up with a negative answer with this. I guess I could define my h value as negative, but I've never had to do that before, and something tells me it would be a bit out of the ordinary to have to start doing that now...I'm not really sure what I'm doing wrong. Any helpful insight would be greatly appreciated! Thanks a mil!

 

[alphysicist]

1. The problem statement:

The coefficient of friction between the 3.00 kg block and surface in the figure below is 0.400. The system starts from rest. What is the speed of the 5.00 kg ball when it has fallen 1.10 m? The picture "below" depicts the 3.0 kg block on the table, attached to a string that is connected to a pulley at the end of the table and is also connected to the hanging 5.0 kg ball.

Homework Equations

PEg: mgh
KE: .5mv^2
Wnc = Change in PEg + Change in KE
Wnc (Distance that non-conservative factor is acting upon system) X (Force that this non-conservative factor is applying)

This last equation is not quite right. One way to write it (for a constant force):

$$W = F d \cos\theta$$

where F and d are the force magnitude and distance, and theta is the angle between the force and displacement vectors.

The Attempt at a Solution

So I assumed that this is a Work Non-Conservative problem because friction is involved. The problem states that the 5.00 kg ball has fallen 1.1 m, and since it is attached by a string to the 3.00 kg block, I assumed that the 3.00 kg block slid along the table 1.1 m. This would be the distance of the non-conservative factor. The force of the non-conservative factor would be equal to the Coefficient of friction X the normal force (in this case, (0.4)(3.0 kg)(9.81m/s^2). This gave me 12.9492.

What is the angle between the frictional force and the displacement? Using that in the equation I gave above shows that this is not quite right.

Next, I assumed that the final potential energy of the 5.0 kg ball would be zero (because i defined the point 1.1 m below it's starting point as PE = 0), therefore the change in PE (PEf-PEi) would be -mgh. Since the ball is starting from rest, there is no initial KE, so I set my equation up as 12.9492 = -mgh + .5mv^2
Obviously, I come up with a negative answer with this.

It's difficult to tell what you might have done here; what numbers did you plug in? (For example, with the way you have written it, the m in the -mgh term is not the same as the m in the .5mv^2 term.)

I'm not certain what you are referring to about a negative answer. If they were asking for the velocity of the 5kg ball you would choose the negative answer once after taking the square root, but here they want the speed. Or were you talking about something else?

 

[drewmacq]

Okay, I completely forgot about the cos(180), which would make the force of friction negative. So instead, now i have:

(.4)(3.0 kg)(9.81 m/s^2) X (1.1 m) X (cos(180)), or -12.9492

So I figured that the only factors I need would be the Initial PEg and the final KE, and therefore I have:

-12.9492 = -mgh + .5mv^2

for my mass in this part, I used 5 kg, since this is the ball that is actually changing in h, so I have:

-12.9492 = -(5 kg)(9.81 m/s^2)(1.1 m) + .5(5 kg)(v^2)

the answer I am coming up with is 4.05 m/s, and this is not correct. I'm not really sure what I'm missing in here...