Solve Trebuchet Efficiency Mystery

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To calculate the energy efficiency of a trebuchet, the user has determined the theoretical yield of gravitational potential energy (10.05 J) and the actual kinetic energy output (43.28 J). The calculated efficiency appears to be 430%, which raises concerns about the accuracy of the results. The discussion suggests using a simpler method for efficiency calculation based on the actual measured range compared to the theoretical range. It emphasizes that efficiency should be calculated as work in divided by work out, indicating that the current approach may have errors in the assumptions or calculations.
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Homework Statement


I've calculated the time, the distance, the height of the counterweight before launch, the mass/weight, and I need to calculate the energy efficiency.
How to go about it?

h = 0.71 m
m = 1.445 kg
g = 9.8 m/s^2
t = 0.7 s
x = 4.44 m

Homework Equations



Eg = mgh
Vix = x/t
y = Viy(t) + (1/2) a(t)^2
Ek = (mv^2)/2

The Attempt at a Solution



THEORETICAL YIELD:
h = 0.71 m
m = 1.445 kg
g = 9.8 m/s^2
Eg = mgh
= (1.445 kg)(9.8 m/s2)(0.71 m)
= 10.05 J

ACTUAL YIELD:
t = 0.7 s
x = 4.44 m
Vix = x/t
= (4.44 m)/(0.7 s)
= 6.34 m/s

y = Viy(t) + (1/2) a(t)^2
0.71 m = Viy(0.7 s) + (1/2) (-9.8 m/s2)(0.7 s)^2
Viy = 4.44 m/s

Vi^2 = (Vx)^2 + (Viy)^2
Vi^ = (6.34 m/s)^2 + (4.44 m/s)^2
Vi= 7.74 m/s

Ek = (1/2) mv^2
= (1/2) (1.445 kg)(7.74 m/s)^2
= 43.28 J

43.28 J / 10.05 J = 430%

I don't think my trebuchet is 430% efficient.
What did I do wrong?
 
Last edited:
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Efficiency is work in/work out
 
An easier way to find trebuchet efficiency (all credit goes to Rom Toms http://www.trebuchet.com/):
Actual measured range in meters /theoretical range= efficiency

theoretical range in meters= 2 * (counterweight in kg * counterweight drop in meters)/ weight of projectile in kg
 
Last edited by a moderator:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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