Two mass pulley system

  • #1

Homework Statement



https://int.erlace.com/uploads/teacher/1533/images/3modifiedatwood(1).jpg [Broken]
Ignore friction and the the rope and the pulley and the rope have negligible mass.
  1. When the hanging block
    gif.latex?M_2.gif
    has descended a distance
    gif.latex?%5CDelta%20h.gif
    , 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
    gif.latex?%5Cmu_k.gif
    between
    gif.latex?M_1.gif
    and the table surface. Find the speed of the masses when
    gif.latex?M_2.gif
    has descended that distance
    gif.latex?%5CDelta%20h.gif
    .

Homework Equations



gif.latex?W%3D-%5CDelta%20U.gif

gif.latex?KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2.gif

gif.latex?U_%7Bg%7D%3Dmgh.gif


The Attempt at a Solution



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

gif.latex?W%3Dmg%5CDelta%20h.gif
gif.latex?%3D%28m_%7B2%7D+m_%7B1%7D%29gh.gif


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
7B1%7D%29gh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv%5E2+%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv%5E2.gif
, 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.

gif.latex?a%3D%5Cfrac%7Bm_%7B2%7Dg-%5Cmu%20m_%7B1%7Dg%7D%7Bm_%7B2%7D+m_%7B1%7D%7D.gif


Thanks.
 
Last edited by a moderator:

Answers and Replies

  • #2
Looks like two pictures didn't post

For part 1, (m1 + m2)gh

For part 3, a = (m2g - μm1g)/(m1 + m2)
 
  • #3
501
66
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 .
 
  • #5
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?
 
  • #6
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 )
 

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