A0.405 kg mass is attached to a spring with a force constant of 26.3 n/m and released from rest a distance of 3.31 cm from the equilibrium position of the spring. calculate the speed of the mass when it is halfway to the equilibrium position.

0.231 m/s

Explanation:

m = mass attached to the spring = 0.405 kg

k = spring constant of spring = 26.3 N/m

x₀ = initial position = 3.31 cm = 0.0331 m

x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m

v₀ = initial speed = 0 m/s

v = final speed = ?

Using conservation of energy

Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy

(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²

m v₀² + k x₀² = m v² + k x²

(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²

v = 0.231 m/s

not really sure but thinking 1.54 mm/us

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures