A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
1 answer:
Answer:
a) 250 N/m
b) 22.4 rad/s , 3.6 Hz , 0.28 sec
c) 0.3125 J
Explanation:
a)
F = force applied on the spring = 7.50 N
x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m
k = spring constant of the spring
Since the force applied causes the spring to stretch
F = k x
7.50 = k (0.03)
k = 250 N/m
b)
m = mass of the particle attached to the spring = 0.500 kg
Angular frequency of motion is given as
= 22.4 rad/s
= frequency
Angular frequency is also given as
= 2 π
22.4 = 2 (3.14) f
= 3.6 Hz
= Time period
Time period is given as
= 0.28 sec
c)
A = amplitude of motion = 5 cm = 0.05 m
Total energy of the spring-block system is given as
U = (0.5) k A²
U = (0.5) (250) (0.05)²
U = 0.3125 J
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