A 2.00kg object is attached to a spring. The force constant of the spring is 20.
ID: 1643494 • Letter: A
Question
A 2.00kg object is attached to a spring. The force constant of the spring is 20.0 N/m. The object is displaced 0.0500m from the equilibrium position. After release, the object exhibits SHM. Determine:
A) Period of the motion: ANS (1.98s)
B) Total energy of the system: ANS (0.0250 J) PLEASE EXPLAIN THIS ONE WELL!
C) Object’s speed when it is 0.0100 meter from equilibrium: ANS (0.154 m/s) PLEASE EXPLAIN THIS ONE WELL!
Physics 1 College level. Please show all work and formula used and that the answer mathes the one given. Thanks!
Explanation / Answer
(A) m = 2 kg
k = 20 N/m
T = 2pi sqrt( m / k )
= 1.99 sec
(b) A = 0.05 m
Total energy = k A^2 / 2 = (20)(0.05^2) / 2
= 0.025 J
(c) total energy = k x^2 /2 + m v^2 /2
0.025 = (20)(0.01^2 / 2) + 2 v^2 /2
v = 0.155 m/s
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