A 0.80 kg mass is attached to a light spring with a force constant of 24.9 N/m a
ID: 2000056 • Letter: A
Question
A 0.80 kg mass is attached to a light spring with a force constant of 24.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass m/s (b) speed of the oscillating mass when the spring is compressed 1.5 cm m/s (c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position m/s (d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m
Explanation / Answer
a)Maximum Speed = A*Omega = 5*10^-2*5.57 = 0.278 m/s
where omega = (k/m)
b)The potential energy in the spring when it is compressed 0.015 m is:
PE = k d² / 2
PE = (24.9 N/m) (0.015 m)² / 2
PE = 0.002801 J
The kinetic energy of the object at that point is:
KE = TE - PE
KE = (0.031125 J) - (0.002801 J)
KE = 0.028304 J
The speed of the object at that point is:
KE = m v² / 2
(0.028304 J) = (0.8 kg) V² / 2
v = 0.266 m/s
c) v = 0.266 m/s
d) Half the maximum speed is:
v = V/2
v = (0.278 m/s) / 2
v = 0.139 m/s
The kinetic energy of the object at that speed is:
KE = m v² / 2
KE = (0.8 kg) (0.139 m/s)² / 2
KE = 0.007728 J
The potential energy in the spring at that time is:
PE = TE - KE
PE = (0.031125 J) - (0.007728 J)
PE = 0.02339 J
The distance from equilibrium at that time is:
PE = k d² / 2
(0.02339J) = (24.9 N/m) d² / 2
d = 0.043 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.