Bob walks 290 m south, then jogs 630 m southwest, then walks 290 m in a direction 26 ∘ east of north.
1 answer:
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Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k