A 76.7 g mass is attached to a horizontal spring with a spring constant of 3.34 N/m and released from rest with an amplitude of 38 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.

0.687 m/s

Explanation:

Given

Mass of object, m = 76.7 g = 0.767 kg

Force constant of the spring, k = 3.34 N/m

Amplitude of the spring, x = 38 cm = 0.38 m

EPE(i) = KE(f) + EPE(f)

1/2kx(i)² = 1/2mv² + 1/2kx(f)²

1/2 * 3.34 * 0.38² = (1/2 * 0.767 * v²) + [1/2 * 3.34 * (0.38/2)²]

1/2 * 3.34 * 0.1444 = (1/2 * 0.767v²) + 1/2 * 3.34 * 0.0361

1/2 * 0.482 = 1/2 * 0.767v² + 1/2 * 0.121

0.241 = 1/2 * 0.767v² + 0.06

1/2 * 0.767v² = 0.181

0.767v² = 0.181 * 2

0.767v² = 0.362

v² = 0.362 / 0.767

v² = 0.472

v = √0.472

v = 0.687 m/s

a.) is the correct answer.

explanation: