Recent content by Diego Cunha

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

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...

Looks like two pictures didn't post For part 1, (m1 + m2)gh For part 3, a = (m2g - μm1g)/(m1 + m2)

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. When the hanging block has descended a distance , how much work has the gravitational force by the Earth done on...