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tobywashere
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Homework Statement
A spring is fired from a launcher angled Θ degrees from the floor. The spring has force constant k and mass M. The target is Δd metres away from the launcher and the target is elevated Δy metres. When the spring is launched, it travels at initial speed v0 towards the target, Θ degrees from the horizontal. The acceleration due to gravity is g = -9.8 m/s2. What is the distance the spring needs to be pulled back, x?
Homework Equations
t = time
x = deformation of spring
Ee = 0.5kx2
Ek = 0.5mv02
Δy = (t)(v0)(sinΘ) + 0.5(g)(t2)
Δd = (v0)(cosΘ)(t)
The Attempt at a Solution
I just need someone to check my calculations. I need to figure out how far to pull back the spring based on a given angle, and a given horizontal distance and vertical distance to the target. There are no specific numbers, I just need a general equation.
The elastic potential energy in the spring is approximately equal to the kinetic energy the spring has after the launch. Air resistance and friction are negligible.
Therefore, 0.5kx2 =0.5mv02
kx2 = mv02
v02 = (kx2) / m
Rearranging Δd = (v0)(cosΘ)(t)
t = (Δd) / (v0 cosΘ)
Substitute t into Δy = (t)(v0)(sinΘ) + 0.5(g)(t2)
Δy = (Δdv0sinΘ/v0cosΘ)(v0)(sinΘ) + 0.5(g)(Δd/v0cosΘ)2
Δy = ΔdtanΘ + (gΔd2)/(2v02cos2Θ)
Δy - ΔdtanΘ = (gΔd2)/(2v02cos2Θ)
2v02cos2Θ = (gΔd2)/(Δy - ΔdtanΘ)
v02 = (gΔd2)/(2Δycos2Θ-2ΔdsinΘcosΘ)
v02 = (gΔd2)/(2Δycos2Θ-Δdsin2Θ)
Substitute v02 = (kx2) / m
(kx2) / m = (gΔd2)/(2Δycos2Θ-Δdsin2Θ)
x2 = (gΔd2m)/(2kΔycos2Θ-Δdksin2Θ)
x = √(gΔd2m)/(2kΔycos2Θ-Δdksin2Θ)
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