anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Since there's no energy lost in the system: PE+KE before = PE + KE after |dw:1351478956312:dw|
anonymous
  • anonymous
Thanks for that reply, I understand that, but how do I set up the equation to calculate the velocity?
anonymous
  • anonymous
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')

Looking for something else?

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