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PLEASE HELP! WILL GIVE BRAINLIEST! A mass of 500 g is attached to a spring with elastic constant coefficient of elasticity = 50.0 N / m. The spring is lengthened by 10 cm and then left free. Calculate the return force exerted by the spring.​

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Answer:

Force = 10N

Explanation:

mass = 500÷1000 = 0.5kg

k. = 50.0 N/m

extension, e = 10÷100 =0.1 m

mg-T = net force

mg-T = ma

but a = -k×e ÷ m from simple harmonic motion

a = -50×0.1 ÷ 0.5

a = -10m/s^2

therefore T = m(g - a)

T = 0.5(10+10)

T = 10N

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