# Trebuchet Efficiency

1. 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.

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

2. Homework Equations

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

3. 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:

Related Introductory Physics Homework Help News on Phys.org
Efficiency is work in/work out

An easier way to find trebuchet efficiency (all credit goes to Rom Toms http://www.trebuchet.com/): [Broken]
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: