Question

A 0.36-kg object connected to a light spring with a force constant of 23.4 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest. (b) Determine the speed of the object when the spring is compressed 1.5 cm. (d) For what value of x does the speed equal one-half the maximum speed?

Answer 1

m = mass = 0.36 kg
k = 23.4 N/m
A = amplitude = 0.04 m

The total energy in the system is:
TE = k A² / 2
TE = (23.4 N/m) (0.04 m)² / 2
TE = 0.01872 J


The potential energy in the spring when it is compressed 0.015 m is:
PE = k d² / 2
TE = (23.4 N/m) (0.015 m)² / 2
TE = 0.0026325 J

The kinetic energy of the object at that point is:
KE = TE - PE
KE = (0.01872 J) - (0.0026325 J)
KE = 0.0160875 J

The speed of the object at that point is:
KE = m v² / 2
(0.0160875 J) = (0.36 kg) V² / 2
v = 0.298 m/s - answer (b)

Half the maximum speed is:
v = V/2
v = (0.322 m/s) / 2
v = 0.161 m/s

The kinetic energy of the object at that speed is:
KE = m v² / 2
KE = (0.36 kg) (0.161 m/s)² / 2
KE = 0.004665 J

The potential energy in the spring at that time is:
PE = TE - KE
PE = (0.01872 J) - (0.004665 J)
PE = 0.01405 J

The distance from equilibrium at that time is:
PE = k d² / 2
(0.01405 J) = (23.4 N/m) d² / 2
d = 0.03465 m - answer (d)