Conservation of energy and spring?
A spring with a spring constant of 5 N/m has a 0.25 kg box attached to one end such that the box is hanging down from the string at rest. The box is then pulled down another 14 cm from its rest position. Calculate the maximum height, maximum speed and the maximum acceleration of the box.
- NCSLv 71 month agoFavorite Answer
ω = √(k/m) = √(5kg/s² / 0.25kg) = 4.47 rad/s
Measured from the new equilibrium position,
y = -A*cos(ωt) = -14cm*cos(4.47t)
for t in seconds.
Where do you want the maximum height measured from?
From the maximum stretch point, the height is
h = 2A = 28 cm ◄
From the weighted equilibrium, it's 14 cm.
V = Aω = 14cm * 4.47rad/s = 63 cm/s ◄
a = Aω² = 280 cm/s² = 2.8 m/s² ◄
Hope this helps!