Energy conservation for objects hanging from a pulley

[agusb1]

Homework Statement

Two objects are connected by a light string passing over a light, frictionless pulley as shown in image. The object of mass m1= 5.00 kg is released from rest at a height h= 4.00 m above the table. Using the isolated system model, (a) determine the speed of the object of mass m2 = 3.00 kg just as the 5.00 kg objects leaves hits the table and (b) find the maximum height above the table to which the 3.00 kg object rises

Relevant Equations

Change in mechanical energy of the system = 0

I have some conceptual questions about this task. In order to get the correct result (I checked the textbook answer) in part (a) I had to assume that the speed for each block is the same at all instants. And that if one block moves down x meters, the other one will move up that same amount of meters. Is this always the case for objects connected through a a light frictionless pulley?

In part (b), when analyzing the maximum height to which the 3kg objects rises, I had to use the following equation to get the correct answer:
m2 * g* 4 m + 0.5 * m2 * (4.42 m/s)^2 = m2 * g* Hmax
(4.42 m/s is the speed obtained when solving part (a) )

The initial situtation we're analyzing is when m1 hits the table. At that point m2 is at h= 4 m. The final situation is the one in which m2 is at the maximum height, that is, its velocity is zero. Now, my question is how can m2 go further up AFTER m1 has touched the table? Shouldn't it stay at 4 meters? Once the block touches the table, it can't keep going down, so it should stop moving. If one block is moving upwards while the other one is at rest, it means their speeds are different. Doesn't that contradict the idea that both blocks have the same speed and acceleration(in magnitude), and that they move the same amount of meters in opposite directions? m1 would already be at h = 0 when m2 is at h =4 meters, and it seems like it can't have a negative height because there's a table stopping it from moving. So how does one object go from h = 4 m to h= 5 m while the other one doesn't move down one meter? Does this imply that once m1 hits the table, it is no longer a part of the pulley system, so those "rules" no longer apply?

 

[vela]

I have some conceptual questions about this task. In order to get the correct result (I checked the textbook answer) in part (a) I had to assume that the speed for each block is the same at all instants. And that if one block moves down x meters, the other one will move up that same amount of meters. Is this always the case for objects connected through a a light frictionless pulley?

Not exactly. It's because the two masses are connected by a taut string that we assume doesn't stretch. Think about two masses connected by a string being pulled in a straight line along a horizontal surface. There's no pulley involved, but each block still moves the same distance because of the string. The pulley simply allows for the displacements to have different directions, e.g. one block moves up while the other one moves down.

The initial situtation we're analyzing is when m1 hits the table. At that point m2 is at h= 4 m. The final situation is the one in which m2 is at the maximum height, that is, its velocity is zero. Now, my question is how can m2 go further up AFTER m1 has touched the table? Shouldn't it stay at 4 meters?

What does your physical intuition tell you would happen here? Can the string exert a force on the second mass to bring it to rest?

 

[256bits]

Now, my question is how can m2 go further up AFTER m1 has touched the table? Shouldn't it stay at 4 meters

m2 has a velocity upwards, doesn't it, when m1 hits the table?

 

[agusb1]

Not exactly. It's because the two masses are connected by a taut string that we assume doesn't stretch. Think about two masses connected by a string being pulled in a straight line along a horizontal surface. There's no pulley involved, but each block still moves the same distance because of the string. The pulley simply allows for the displacements to have different directions, e.g. one block moves up while the other one moves down.What does your physical intuition tell you would happen here? Can the string exert a force on the second mass to bring it to rest?

I would think the weight of mass 1 is pulling the other block up? The table also exerts a force on mass 1. Is that the force that makes it lose velocity and reach v = 0? Will the isolated system model stop being valid once mass 1 has touched the table?

 

[kuruman]

I would think the weight of mass 1 is pulling the other block up?

Not quite. The left side of the string is pulling the left mass up and the Earth is pulling the left mass down. The same happens on the right side. Now the force (tension) with which the string pulls each mass up is the same, but the Earth pull is larger on the more massive object. Thus, the net force on the more massive object is down while the net force on the less massive object is up.

The table also exerts a force on mass 1. Is that the force that makes it lose velocity and reach v = 0?

Yes.

Will the isolated system model stop being valid once mass 1 has touched the table?

Yes. Once part of the system interacts with the table, it is no longer isolated unless you consider the table part of the isolated system. I have a question for you on this. Before the mass hits the table, what do you think are the components of your isolated system? Specifically, do you include the Earth as part of the isolated system or not?

 

[agusb1]

Yes. Once part of the system interacts with the table, it is no longer isolated unless you consider the table part of the isolated system. I have a question for you on this. Before the mass hits the table, what do you think are the components of your isolated system? Specifically, do you include the Earth as part of the isolated system or not?

In part (a) I considered the components of my system to be the two blocks and the Earth, but in part (b) I considered only mass 2 and the Earth. I treated it as an isolated system, however. I wrote that change in kinetic energy = - (change in potential energy). So, why did this equation work? Is the system consisting of mass 2 and the Earth isolated while the system consisting of mass 1, mass 2 and the Earth is not isolated?

 

[kuruman]

In part (a) I considered the components of my system to be the two blocks and the Earth, but in part (b) I considered only mass 2 and the Earth. I treated it as an isolated system, however. I wrote that change in kinetic energy = - (change in potential energy). So, why did this equation work? Is the system consisting of mass 2 and the Earth isolated while the system consisting of mass 1, mass 2 and the Earth is not isolated?

In part (a) the isolated system is the two masses, the rope and the Earth. In part (b) both masses are not isolated from the Earth but masses 1 and 2 can be considered isolated from each other as long as the rope is slack (you can pull with a rope but you cannot push with a rope). So as long as the rope is slack, there are two systems involving the masses that can be considered isolated: Mass 1 + table + Earth and mass 2 + Earth. When mass 2 comes back down and the rope is again under tension, we are back to the isolated system in part (a).