# A system of two paint buckets

Conservation of Energy

## Homework Statement

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0 kg bucket 2.00 m above the floor. Using the principle of conservation of energy to find the the speed with which this bucket hits the floor. You can ignore friction and the mass of the pulley. The other bucket has the mass 4.0 kg.

NOTE: it is essentially a pulley with the 4.0kg bucket on the floor and the 12.0kg bucket 2m off the floor.

K1+U1=K2+U2

## The Attempt at a Solution

U1=0
K1= (1/2)(4)(v^2)
U2=(12)(9.8)(2)
K2=(1/2)(12)(v^2)

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The equation you have is right, but you have to think about the system.
K1 would be zero, because at the beginning, nothing is moving, and U1 will be the potential of the 12 kg bucket if you set the ground as reference point.
Then as the 12 kg bucket falls and hits the ground, the 4kg bucket will move to 2m, becoming your potential at the end system while the 12kg bucket hits the floor with the unknown velocity you are trying to find.
So it is better if your equation was written as
Ko+Uo=Kf+Uf

Hope this helps

wait so, in that case

Ko+ Uo = Kf + Uf

0 + (12)(9.8)(2) = (1/2)(4)(v^2) + (4)(9.8)(2)

i get a v that is twice the answer. the answer is 4.4

you forgot about the 12kg bucket which is also moving.
so your equation should really be

Ko+ Uo = Kf + Uf

0 + (12)(9.8)(2) = (1/2)(4)(v^2) +(1/2)(12)(v^2)+ (4)(9.8)(2)

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thanks so much!!
only 9283923 questions left =/