A 120 g block hooked to a horizontal spring with spring constant 5 N/m is 4...

Question

Question

A 120 g block hooked to a horizontal spring with spring constant 5 N/m is 4 cm from the equilibrium position and moving at 2.

science physics

Answer 1

Answer #1

m = 0.12kg

K = 5N/m

x = 0.04m

V = 2.1m/s

A.) Total energy E = 1/2 kx^2 + 1/2 mv^2

=) E = 0.5( 5×0.04×0.04 + 0.12×2.1×2.1)

=) E = 0.2686 Joule

B.) Since maximum potential energy U = E = 1/2 kA^2

=) A = (2U/k)^0.5 = (2×0.2686/5)^0.5 = 0.32778m

=) Amplitude A = 0.32m

Answer 2

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