# Using Energy Concept

1. Oct 15, 2006

### BunDa4Th

Three objects with masses m1 = 4.0 kg, m2 = 10.0 kg, and m3 = 15.0 kg, respectively, are attached by strings over frictionless pulleys, as indicated in Figure P5.32. The horizontal surface is frictionless and the system is released from rest. Using energy concepts, find the speed of m3 after it moves down a distance of 2.5 m.

Okay, I am lost at this. At first I thought this was a simple problem using
V^2 = 2adeltaX which apparently is not, if it is then i am doing something wrong. I read on the conservative energy in the book I am using and understand a bit of it but not sure where to start.

I seriously dont know where to start or what i am doing right now. I seem to enter number randomly sometimes.

Last edited by a moderator: Apr 18, 2017
2. Oct 15, 2006

### Repetit

First of all you should realize that all the masses have the same speed because the wire is unstretchable. Next, try to imagine what happens when the system is released. Because M3 is heavier than M1, and because the horizontal surface is frictionless, M3 will fall.

Now, when I solve problems like this I think to myself: "What puts a limit to the speed of M3". The anwer is the kinetic energy gained by M2, and the potential energy gained by M1 when it moves up. So you find the total energy available (which would be M3*g*2.5), and subtract the energies due to the limiting factors. This gives you the total energy available to the kinetic energy of M3.

3. Oct 15, 2006

### BunDa4Th

the protential energy for m1 would be 98

the total energy available is 367.5

I am not sure how to find the kinetic energy of m2 since KE = 1/2mv^2 not knowing what v is how would i go about solving this?

4. Oct 15, 2006

### BunDa4Th

can anyone explain how this problem work in simple terms?

5. Oct 16, 2006

### Repetit

Just write the complete equation:

$$\frac{1}{2}M_3 v^2 = M_3 g \Delta h - \frac{1}{2}M_2 v^2 - M_1 g \Delta h$$

Where the term om the left is the kinetic energy of M3 (the one you want to calculate), the first term on the right is the total energy available (with $$\Delta h=2.5m$$, the second term is the kinetic energy of M2 and the last term is the potential energy increase of M1. So you get an equation with only one unknown.

You don't have to know v of M2, because it is the same v as for M3.

6. Oct 16, 2006

### Repetit

The concept would be the same even if the horizontal surface was not frictionless. Then you would just subtract the energy delivered to the surface due to friction as well, and this energy would be the the frictional force times the distance travelled (2.5m).

7. Oct 16, 2006

### BunDa4Th

thanks for explaining this. I now know what mistake i made. I was trying to solve everything on its own instead of making it into one equation.

thanks again.

8. Oct 16, 2006

### BunDa4Th

okay i did this problem and it said i was incorrect.

(1/2)15v^2 = (15)(9.8)(2.5) - 1/2(10)v^2 - (4)(9.8)(2.5)

7.5v^2 = 367.5 - 5v^2 - 98

12.5v^2 = 269.5

v^2 = 21.56

v = 4.64

It said that it was within 10% of the correct answer. Can anyone point out the mistake i made?

9. Oct 17, 2006

### BunDa4Th

well, i gave this problem another try thinking maybe i input "g" wrong for one of them, which still gave me the wrong answer. I am really confuse what i am doing wrong here.

10. Oct 17, 2006

### BunDa4Th

anyone? I tried using this

KE_i + PE_i = KE_f + PE_f which i think is the same thing as the above formula but for some reason when i did this equation i got a whole completely different number.

11. Oct 19, 2006

### Repetit

Hey... I think you should subtract the kinetic energy gained by M1 as well. It does indeed have a kinetic energy when M3 has fallen a distance of 2.5m... Sorry, I forgot about this :-)

12. Oct 19, 2006

### BunDa4Th

Yea, i was able to figure this out after I realize something was missing after about 5 tries. Thanks for setting me to the right track.