Calculating Work and Kinetic Energy in a Two Mass Pulley System

[Diego Cunha]

Homework Statement

https://int.erlace.com/uploads/teacher/1533/images/3modifiedatwood(1).jpg
Ignore friction and the the rope and the pulley and the rope have negligible mass.

1. When the hanging block
   
   has descended a distance
   
   , how much work has the gravitational force by the Earth done on the system?
2. That work went into the kinetic energy of the masses. What fraction of that kinetic energy went into the motion of the hanging block?
3. Suppose now there's a coefficient of kinetic friction
   
   between
   
   and the table surface. Find the speed of the masses when
   
   has descended that distance
   
   .

Homework Equations

The Attempt at a Solution

1. I figured the work done was equal to change in potential energy, so

2. I'm not sure how to go about this one. I tried to take what I found for the work done and relate it the kinetic energy of the masses, so something like

, but I don't know if this would help, or if it is right at all.

3. Not sure. I derived an expression for acceleration of the system, and again I don't know if it would help.

Thanks.

 

[Diego Cunha]

Looks like two pictures didn't post

For part 1, (m₁ + m₂)gh

For part 3, a = (m₂g - μm₁g)/(m₁ + m₂)

 

[Qwertywerty]

Your solution :

(1.) is wrong .
Does gravity do work on both blocks ?

*Remember - Work done is ∫F.dx

(2.) - What is acceleration of each block ? So what is velocity after moving a distance Δh ?

(3.) - Repeat as previous part .

I hope this helps .

 

[haruspex]

For part 3, a = (m₂g - μm₁g)/(m₁ + m₂)

You got that part right.

 

[Diego Cunha]

Your solution :

(1.) is wrong .
Does gravity do work on both blocks ?

*Remember - Work done is ∫F.dx

(2.) - What is acceleration of each block ? So what is velocity after moving a distance Δh ?

(3.) - Repeat as previous part .

I hope this helps .

Yes, this helps. Gravity would only do work on m₂ so the work done would be equal to m₂gh. The acceleration of the blocks are the same, so each would have an an acceleration a = (m₂g)/(m₁+m₂ ), without friction. For the m₂, the velocity would be (m₂v²)/2 = m₂gh, so v = √(2gh). Does this tell me anything in terms of how much kinetic energy went into moving m₂, would it be all of it?

 

[Diego Cunha]

Yes, this helps. Gravity would only do work on m₂ so the work done would be equal to m₂gh. The acceleration of the blocks are the same, so each would have an an acceleration a = (m₂g)/(m₁+m₂ ), without friction. For the m₂, the velocity would be (m₂v²)/2 = m₂gh, so v = √(2gh). Does this tell me anything in terms of how much kinetic energy went into moving m₂, would it be all of it?

I think my expression for velocity was wrong, since I didn't account for m₁, it would be v=√2h(m₂g)/(m₁+m₂ )