# Conservation of engery of a pulley system

1. Mar 12, 2010

### jwxie

1. The problem statement, all variables and given/known data

Two objects, m1 = 4.50 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 4.50 kg is released from rest, h = 3.00 m above the ground.

(a) Using the isolated system model, determine the speed of the 3.00 kg object just as the 4.50 kg object hits the ground.
(ans: 4.43)

(b) Find the maximum height to which the 3.00 kg object rises.
(ans: 5)

2. Relevant equations

$$\Delta K$$ = - [tex]\Delta Ug[/text]
mgy - mgyf = - [tex]\Delta Ug[/text]
KEi + PEi = KEf + PEf

3. The attempt at a solution

I know I can solve problem a differently. Suppose I solve each mass separately using law of conservation of energy, I can find Vf for mass 2 separately.

Since the system is connected by the same uniform massless string, so the accerlation must the same. But I did not assume their Vf will be the same (the impact of m1 as it hits the ground, and at that instant the Vf of m2 raising to 4 meter).

What I want to know is, how do you prove that the final velocity of m1 and m2 at that same instant is the same using the isolated system model KEf + PEf = KEi + PEi ?

I have the following data on my hands

for m2, where m= m2, h = 4
_________________________
| | KE | PE |
| i | 0 | 0 |
| f | 1/2mv^2 | m2h |

for m1 where m = m1, h = 4
| | KE | PE |
| i | 1/2mv^2 | mgh |
| f | 0 | 0 |

Last edited by a moderator: Apr 17, 2017
2. Mar 12, 2010

### rl.bhat

Since the string is unstretchable, if m1 moves a small distance Δx in time Δt, m2 also moves the same distance in same time interval. So vf of m1 is the same as vf of m2.

3. Mar 12, 2010

### jwxie

Hi, thank you.