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), where A is the amplitude (0.080 m), ω is the angular frequency (√(k/m) = √(190/1.6)), and t is the time (0.40 s).

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