A 0.440 kg mass is attached to a light spring with a force constant of 28.90 N/m
ID: 1549752 • Letter: A
Question
A 0.440 kg mass is attached to a light spring with a force constant of 28.90 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.00 cm and released from rest, determine the following. maximum speed of the oscillating mass 0.405 m/s speed of the oscillating mass when the spring is compressed 1.50 cm 0.387 m/s speed of the oscillating mass as it passes the point 1.50 cm from the equilibrium position 0.387 m/s value of x at which the speed of the oscillating man is equal to one-half the maximum value 0.433Explanation / Answer
d)
xmax = maximum initial expansion of spring = 5 cm = 0.05 m
k = spring constant = 28.90 N/m
v = speed = Vmax/2 = 0.405/2
x = ?
m = mass attached = 0.440 kg
using conservation of energy
maximum spring energy = kinetic energy + spring energy
(0.5) k xmax2 = (0.5) m v2 + (0.5) k x2
(0.5) (28.9) (0.05)2 = (0.5) (0.44) (0.405/2)2 + (0.5) (28.9) x2
x = 0.0433 m
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