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A 1.0 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

User KarlM
by
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1 Answer

5 votes

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


User Federico Taschin
by
6.3k points