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Finding mass in a pulley system

  1. Jun 29, 2013 #1
    1. A simple Atwood’s machine uses a massless pulley and two masses m1 and m2. Starting from rest, the speed of the two masses is 2.2 m/s at the end of 2.9 s. At that time, the kinetic energy of the system is 85 J and each mass has moved a distance of 3.19 m

    (a) Find the value of heavier mass. The acceleration due to gravity is 9.81 m/s^2.


    (b) Find the value of lighter mass.

    2. Relevant equations
    KE=1/2mv^2
    F=ma

    3. The attempt at a solution

    well, it says that i have the kinetic energy so it would be something like this:

    85J= 1/2m1(2.2m/s)^2 + 1/2m2(2.2m/s)^2

    after this i get stuck, they are asking me to find the mass, and the only other equation that i think may fit would be F=ma but then, I don't know how
     
  2. jcsd
  3. Jun 29, 2013 #2
    You are right thinking about F = ma. Take m1. What force acts on? What is it acceleration? Ditto for m2.
     
  4. Jun 29, 2013 #3
    would it be T? if so then, T-m1a= m1g?
     
  5. Jun 29, 2013 #4
    What about m2?
     
  6. Jun 29, 2013 #5
    wouldn't it be the same? because they are suspended pulleys they have tension on a cord, and gravity acting on them?
     
  7. Jun 29, 2013 #6
    They have the same acceleration but not the same force.
     
  8. Jun 29, 2013 #7
    so, it would be:

    T-m1g=m1a
    T+m2g=m1a ?
     
  9. Jun 29, 2013 #8
    According to that equation both gravity and tension are acting in the same direction on the 2nd mass. Is that correct? Does the tension push it downwards or pull it upwards?
     
  10. Jun 29, 2013 #9
    ooh! then, m2g= m2a ?
     
  11. Jun 29, 2013 #10
    No no - that would be saying that tension doesn't exist for m2! You need to look at the signs in your 2nd equation.
     
  12. Jun 29, 2013 #11
    T-m1g=m1a
    T+m2g=m2a i just change the sign to a negative?
     
  13. Jun 29, 2013 #12
    T-m1g=m1a
    T-m2g=-m2a

    maybe like this?
     
  14. Jun 30, 2013 #13
    Yes, that is fine.
     
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