Why should I use conservation of energy?

AI Thread Summary
Using conservation of energy simplifies the analysis of the system involving two masses and a pulley, as demonstrated in the discussion. The initial approach using free body diagrams led to incorrect calculations, primarily due to errors in substituting values. The correct method involves recognizing that the total mass affecting acceleration includes both masses and half the pulley's mass. A suggestion was made to delay plugging in numerical values until the final stages of the calculations to avoid mistakes. Ultimately, using conservation of energy can yield the correct time of 2.06 seconds for the falling mass.
zewei1988
Messages
21
Reaction score
0

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
 
Last edited by a moderator:
Physics news on Phys.org
You should get the same answer either way. Post your calculations so we can see where your attempt went wrong.
 
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).
 
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).
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top