Conservation of Energy: Finding the Speed of a Falling Ball

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SUMMARY

The discussion focuses on calculating the speed of a falling ball using the conservation of energy principle. The system involves a block of mass m1 = 3.00 kg and a falling ball of mass m2 = 5.00 kg, with a coefficient of friction μk = 0.400. The equation used is m2gh - fh = 1/2(m1 + m2)v², leading to a calculated speed of 4.77 m/s after falling a distance of h = 1.50 m. The user expresses uncertainty regarding the correctness of this result compared to an answer sheet.

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sireh
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Homework Statement


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The coefficient of friction between the block of mass m1 = 3.00kg and the surface in Figure P7.22 is μk = 0.400. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance of h = 1.50m?

Homework Equations


This is in the chapter on conservation of energy, so
Ui + Ki + ΔEmech = Uf + Kf

The Attempt at a Solution


Using downward direction as positive so g is positive

m2gh - fh = 1/2(m1 + m2)v2
(5.00kg)(9.8m/s2)(1.5m) - (-11.76N)(1.5m) = 1/2(3.00kg + 5.00kg)v2
91.14Nm = 4kg * v2
22.785Nm2/s2 = v2
4.77m/s = v

That's not the answer that I'm seeing on the answer sheet. I'm not sure what I did wrong.
 
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sireh said:
m2gh - fh = 1/2(m1 + m2)v2
(5.00kg)(9.8m/s2)(1.5m) - (-11.76N)(1.5m) = 1/2(3.00kg + 5.00kg)v2
Think conceptually and decide if you want to have two minus signs here.
 

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