Question

Two objects are connected by a string that passes over a frictionless pulley, as shown, where m1g c) a2=g d) a2a1

Answer 1

http://s3.amazonaws.com/answer-board-image/c15feec4-4370-493e-9e43-9d8a4d4ede05.gif

Answer 2

@CliffSedge

Answer 3

@ivanmlerner

Answer 4

In physics, before anything else, draw a picture!
Do you know what a FBD is?

Answer 5

yep, i actually drew one on my sheet lol

Answer 6

For this problem, I'm assuming the system starts from rest? If there was an initial external force, it would change matters.

Answer 7

yea, from rest.

Answer 8

Consider that both masses are connected and move as a single object.

Answer 9

So, for any body, the aceleration of gravity is the same. The presence of the mass m1 interacts in wich way with the mass m2? Deacelerating it or acelerating it?
Why would an initial external force change things?

Answer 10

m1, puts an extra mass on m2, which means m2 does not accelerate as much. m1's acceleration depends on m2's acceleration. m1 is a negative acceleration because it moves up while m2 has a positive acceleration and moves down. I assigned down as positive in my FBD.

Answer 11

@CliffSedge the last question is for you.

Answer 12

The external force would add or subtract from velocity, changing acceleration along with it.

Answer 13

I can already conclude that B, C, and D are wrong

Answer 14

You can also conclude that E is wrong too, because the two masses are connected by a (I'm assuming, non-elastic) string. That means that the magnitude of acceleration of both objects must be the same.

Answer 15

but isn't m2 greater? doesn't that mean a2 must also be greater?

Answer 16

Not if they are connected into a single moving system . . .

Answer 17

grater than a1 only

Answer 18

but cliff, look at the diagram, isn't it a single moving system, technically?

Answer 19

The greater mass of m2 means that (starting from rest) the system will move in the direction of m2, but the question asks about the magnitude of the acceleration, not the direction.

Answer 20

Yes, it is a single moving system - which means that there is a single acceleration for the entire system. a1=a2.

Answer 21

Still dont get why an external force would change anything.

Answer 22

An external force (besides just gravity) could pull down on m1, making it move in the other direction, but we are assuming it starts from rest and the only forces are the weights of the two masses.

Answer 23

Ohhhh ok cliff...so wouldn't it be either a or b depending on the numerical mass of m2?

Answer 24

Yes!

Answer 25

And you know which mass is greater, so . .

Answer 26

Yes, but you said thtat an initial external force would change things. So during the time that we are concerned there would be no external force, even if the system had an initial velocity it would not change the aceleration.

Answer 27

a2<g? because g is always 9.8m/s^2, and if m2 is large enough, then can't a2>g at one point? or will a2 always be approaching g?

Answer 28

Well, actually, again, direction isn't important, so it doesn't depend on the difference between m1 and m2. It's only asking about magnitude (size) of the accel.

Answer 29

@ivanmlerner yes, that is true as long as the initial external force ceased to act, then we would be back at the accelerations we have now (starting from rest). But if the additional external force continued to act . . .

[but anyway, that is irrelevant here]

Answer 30

Here's what's important: a1=a2 because the masses are connected. a2 would =g if it was by itself, but it is being affected by m1.

Answer 31

Oh, ok I misinterpreted what you said.

Answer 32

oh all right, so since m1 can never be 0, a2 will always be approaching g, but it can never equal nor be greater than g. right?

Answer 33

One thing you seem to be confused about since you mentioned that a2 would surpass g, is that the aceleration of gravity can't be surpassed without a downward force. You said that if m2 was large enough, then it could do it, but it is impossible.

Answer 34

Well, technically (mathematically) m1 could be zero, but if it was, it would no longer have physical significance and wouldn't need to be mentioned, so yes, consider m1>0.

This is best considered with Newton's second law. The acceleration is determined by the *net* force.

Answer 35

@ivanmlerner is referring to this statement, "...and if m2 is large enough, then can't a2>g at one point? or will a2 always be approaching g?"
a2 can approach g, but not exceed it.

Answer 36

All right. I get it now, thanks! My physics test on all of this is on thursday, but i wanted to clear up any misconceptions now just in case haha. I'll keep practicing and if I have issues I'll let one of you know! Is that all right?

Answer 37

Sure. Stick with the basics. Understand kinematics well, conservation of energy well, Newton's laws, conservation of momentum, etc.
Use formulas to guide your thinking.
And always draw a picture!

Answer 38

Well, I won't be here, because its almost midnight here.

Answer 39

I'm about to go to bed too . . .

Answer 40

yea it's 11pm here right now haha. And all right Cliff, I'll be sure to draw pictures. It's tuesday guys, I mean I'm going to practice on wednesday lol.