Why should I use conservation of energy?

In summary, the conversation discusses a physics problem involving a system with a mass of 23.00 kg, 13.60 kg, and 5.00 kg, as well as a pulley with a radius of 0.13 m. The problem is solved using both free body diagrams and conservation of energy, but the calculations do not match up. The expert suggests checking for errors in the calculations and recommends not plugging in numbers until the end to avoid mistakes.
  • #1
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Homework Statement


Consider the system shown in the figure with m1 = 23.00 kg, m2 = 13.60 kg, R = 0.13 m, and the mass of the uniform pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 5.00 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.


Homework Equations


KEi + Ugi = KEf + KEf(rotational) + Ugf


The Attempt at a Solution


Initially I keep trying to solve the problem by using free body diagrams and finding the acceleration. But all my answers were wrong.
However, I saw someone posted this question on yahoo answers and the solution he got was by using conservation of energy.

http://answers.yahoo.com/question/index?qid=20090423164312AAs5DN8"

I do not see the difference in using the conservation of energy method and my method. Can someone enlighten me? This topic is killing me. haha
 
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  • #2
You should get the same answer either way. Post your calculations so we can see where your attempt went wrong.
 
  • #3
vela said:
You should get the same answer either way. Post your calculations so we can see where your attempt went wrong.

Ok, here are my workings:


Eqn 1:
23(9.8) - T1 = 23a
T1 = 225.4 - 23a

Eqn2:
T2 = 13.6a + 133

Eqn3:
0.13T1 - 0.13T2 = (0.5 * 5 * 0.132) * (a / 0.13)
T1 - T2 = 2.5a

Sub Eqn2 and Eqn1 into Eqn3:
92.4 = 158.5a
a = 0.583 m/s2

5 = 0.5 * 0.583 * t2
t = 4.14s


The correct answer is 2.06s (about half the time I calculated).
 
  • #4
zewei1988 said:
Ok, here are my workings:


Eqn 1:
23(9.8) - T1 = 23a
T1 = 225.4 - 23a

Eqn2:
T2 = 13.6a + 133

Eqn3:
0.13T1 - 0.13T2 = (0.5 * 5 * 0.132) * (a / 0.13)
T1 - T2 = 2.5a
Good up to here, but something went wrong when you plugged the numbers in.
Sub Eqn2 and Eqn1 into Eqn3:
92.4 = 158.5a
The coefficient of a should be the sum of the two masses and half the pulley's mass, 13.60+23.00+0.5(5.00)=39.10 kg, not 158.5 kg.

One suggestion I have is to not plug any numbers in until the end. You probably could have avoided the error in this particular case.
a = 0.583 m/s2

5 = 0.5 * 0.583 * t2
t = 4.14s


The correct answer is 2.06s (about half the time I calculated).
 

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