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A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0 Nm-1. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.

Sagot :

LammettHashidn

The work required to stretch the spring by a displacement of 5.2 cm = 0.052 m is

1/2 (7.0 N/m) (0.052 m)² ≈ 0.0095 J

which is stored as potential energy in the spring. When the mass is released, as the spring relaxes, this potential energy is gradually converted into more and more kinetic energy.

Let x be the displacement (relative to the equilibrium position, with 0 < x < 0.052 m) at which this energy is split evenly between potential energy P and kinetic energy K, so that

P + K = 2P = 0.0095 J

or

P ≈ 0.0047 J

At this displacement, the spring is storing

P = 1/2 (7.0 N/m) x²

of potential energy. Solve for x :

1/2 (7.0 N/m) x² ≈ 0.0047 J

x² ≈ (0.0047 J) / (1/2 (7.0 N/m))

x ≈ √((0.0047 J) / (1/2 (7.0 N/m)))

x ≈ 0.037 m = 3.7 cm

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