Question

The coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure below is μk = 0.280. 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 = 1.10 m?

So i drew force diagrams for both objects.
M2A=M2G-T
M1A=T-M1GUK
I substituted T from the top into the bottom and solved for A, I found that A=5.096 m/s^2 and then i found that T =23.52 N
The net force is this
NetF=M2*G-M1*Uk*G
NetF=40.768 N
I also know W=delta K + delta PE
W=F*D=40.768
W=delta K - MGH
Delta K = W + M2GH

Answer 1

Heres a picture
http://www.webassign.net/serpse8/8-p-022.gif

Answer 2

Delta K =(1/2)MV^2
V=4.96 m/s
But this is wrong, can someone explain what I am doing wrong?

Answer 3

You needn't go into the net force and work; it simplifies some things, but not this.

Since you know the acceleration of the system, go back to one of your five basic kinematic formulas:

\[v _{f}=v _{i }+2ad\]

Answer 4

oops - square the two velocities.

Answer 5

Thanks, I can believe it was that simple. But do you know why the way I attempted it didnt work. I took the net force over the distance to get the work. The change in PE was the mass 2 *G*H. So since I am able to calculate delta K, shouldnt I be able to get the correct velocity from that?

Answer 6

You assumed conservative forces - friction was the fly in the ointment.

Answer 7

Oh i see, so the equation W=delta K + delta PE only applies for conservative forces?

Answer 8

Yes.

Answer 9

Thank you so much, this problem was bothering me for far to long.

Answer 10

Well, you're likely to remember it, then!

Answer 11

Yeah, i guess its the best way to learn, thanks again!