A1.0 kg weight suspended from a spring is pulled to 0.50 m below its equilibrium point. if the spring has a spring constant (k) of 50.0 n/m, at what rate will the mass accelerate when it is released? a) 25 m/s² b) 6.3 m/s² c) 100 m/s² d) 3.7 m/s²

m = mass of the weight suspended = 1 kg

x₀= stretch in the spring at equilibrium = ?

k = spring constant of the spring = 50.0 N/m

using equilibrium of force ,

spring force on the weight = force of gravity on the weight

k x₀ = mg

50 x₀ = (1 x 9.8)

x₀ = 0.196 m

x = distance by which the weight is pulled below equilibrium = 0.50 m

a = acceleration of the weight

F = spring force on the weight in upward direction ,

force equation for the motion of weight is given as

F - mg = ma

k (x + x₀) - mg = ma

(50)(0.50 + 0.196) - (1) (9.8) = 1 a

a = 25 m/s²

the results is a decrease in air temperature with an increase of altitude.

explanation:

0.19 m/s and 0.45 m/s