Conservation of Energy/Gravitational Potential Energy

AI Thread Summary
The discussion revolves around a physics problem involving two buckets connected by a pulley, where one bucket is raised 2.00 m above the ground and has a mass of 4.0 kg. The conservation of energy principle is applied to determine the speed of the second bucket when it hits the ground. Initially, the user calculated the speed as 5.11 m/s, but the textbook states it should be 4.4 m/s. After realizing both buckets possess kinetic energy, the user adjusted their calculations accordingly, leading to the correct answer. The final solution was confirmed to be accurate after the adjustment.
Dougggggg
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Homework Statement


There are 2 buckets hanging from a single pulley with rope (not considering the mass) that connects them. One is on the ground and has a mass of 4.0 kg, the second is 2.00 m above the ground when released. Using conservation of energy, find the speed of the second bucket when it hits the ground.


Homework Equations


K1+U1-Wother=K2+U2

Ug=mgh

K=\frac{1}{2}mv2


The Attempt at a Solution


I put K1 and U2 equal to 0 and made Wother equal to m1gh. Then just solved for v by multiplying by 2/m2 and square rooting.

PHY2110755.jpg


I got 5.11 m/s but the book says 4.4 m/s. Anyone know where I messed up or what I'm not accounting for?
 
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Hi Dougggggg! :smile:

Both buckets have KE. :wink:
 
tiny-tim said:
Hi Dougggggg! :smile:

Both buckets have KE. :wink:

So I would set it equal to a second K on the right side of the equation? I will try it out and edit when I have solved.

Awesome, worked perfectly, thanks.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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