kachexy
contestada

A 1.3 kg mass is attached to a spring with a force constant of 52 N/m. If the mass is released with a speed of 0.28m/s at a distance of 8.1 cm from the equilibrium position of the spring, what is its speed when it is halfway to the equilibrium position?

Respuesta :

Answer:

0.52 m/s

Explanation:

Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

½ m v₁² + ½ k x₁² = ½ m v₂² + ½ k x₂²

m v₁² + k x₁² = m v₂² + k x₂²

Given:

m = 1.3 kg

k = 52 N/m

v₁ = 0.28 m/s

x₁ = 0.081 m

x₂ = 0.081 m / 2 = 0.0405 m

(1.3) (0.28)² + (52) (0.081)² = (1.3) v² + (52) (0.0405)²

v = 0.52

The velocity halfway to the equilibrium position is 0.52 m/s.