A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m.The block is pulled from its equilibrium position at x = 0.00 m to a displacement
X = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.What is the velocity of the block at time t = 0.40 s?
A) 0.82 m/s
B) -0.82 m/s
C) 0.30 m/s
D) -0.30 m/s
E) 0.00 m/s

Question

Quizplus · Accepted Answer

The velocity of the block at time t = 0.40 s is 0.82 m/s. This is calculated using the formula for velocity in simple harmonic motion: v(t) = -Aωsin(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant. Since the block is released from rest at maximum displacement, φ = 0. Substituting the given values, the velocity is found to be 0.82 m/s.

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