A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A = 0.1 m. What is the kinetic energy of the mass when its position is at x = 0.05 m? g
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Ответ:
Kinetic energy at 0.05 m is 0.037 J
Explanation:
Given:
Mass, m = 2 kg
Spring constant, k = 10 N/m
Amplitude, A = 0.1 m
Angular frequency, ω = √k/m
Substitute the suitable values in the above equation.
ω = 2.24 s⁻¹
Simple harmonic equation is represent by the equation:
x = A cos ωt
Substitute 0.05 m for x, 0.1 m for A and 2.24 s⁻¹ for ω in the above equation.
t = 0.47 s
Kinetic energy at x = 0.05 is determine by the relation:
Substitute the suitable values in the above equation.
E = 0.037 J
Ответ:
it keeps runen form the cops dk
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are we talk about thomas the choco train