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Glenboro
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Homework Statement
A ball bearing of mass m = 50.0 g, is sitting on a vertical spring whose force constant is 120.0 N/m. The initial position of the spring is at y = 0 m[/B]
a) The spring is compressed downward a distance x = 0.200 m. From the compressed position, how high will the ball bearing rise? How high does the ball bearing rise above the equilibrium position at y = 0 m?
b) What is the kinetic energy of the ball at the moment it is released from the spring? (assume all energy converted into kinetic energy)
c) What is the maximum speed of the ball?
Homework Equations
a) PEs = (0.5)(K)(x)2, h = PE/mg
c) root[(2)(KE)/m]
The Attempt at a Solution
a)PEs = (0.5)(120.0 N/m)(0.200m)2
= 2.4J
h = PE/mg
h = 2.4J/(0.05kg(9.8))
h = 4.89m
h = 4.89m - 0.200m (the distance of spring compression)
h = 4.69 m
b) KE = 2.4J - mgh
KE = 2.4 J -(0.050kg)(9.8)(0.200m)
KE = 2.3J
c) v = root [2(2.3)/(0.050kg)]
v = 9.60 m/s