Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DonReynolds

  • 3 years ago

A system of two block with one frictionless pulley is release. Block one mass 3.00kg sitting on ground, block 2 mass 5.00 kg with a height of 4.00 meter. What is the velocity (speed) of the 5.0kg block in m/s when it strikes the ground. We have been using the conservation of momentum and Kinetic Energy (KE) and Potential Energy (PEg) formulas. ow do I set this up to solve?

  • This Question is Open
  1. imron07
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Since there's no energy lost in the system: PE+KE before = PE + KE after |dw:1351478956312:dw|

  2. DonReynolds
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks for that reply, I understand that, but how do I set up the equation to calculate the velocity?

  3. imron07
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Before: m1gh1+m2gh2+1/2 mv1^2+1/2 mv2^2=0+5g*4+0+0 After: m1gh'1+m2gh'2+1/2 mv'1^2+1/2 mv'2^2=3g*4+0+1/2x3xv'^2+1/2x5xv'^2 (v'1=v'=v')

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy