Question

A single mass m₁ = 4.4 kg hangs from a spring in a motionless elevator. The spring is extended x = 14 cm from its unstretched length.

Now, three masses m₁ = 4.4 kg, m₂ = 13.2 kg and m₃ = 8.8 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above.

1) What is the force the top spring exerts on the top mass?

2) What is the distance the upper spring is extended from its unstretched length?

Answer 1

Given,

m1 = 4.4 kg ; m2 = 13.2 kg ; m3 = 8.8 kg

1)Since the top spring is supporting all three masses, so the force that it is exerting will be equal to the gravitational pull of all three masses.

Ftop = (m1 + m2 + m3) g = (4.4 + 13.2 + 8.8) x 9.81 = 259 N

Hence, Ftop = 259 N

2) We know from Hook's law that,

k = F/x = m1 g / x = 4.4 x 9.8 / 0.14 = 308 N/m

x' = F/k = Mtot x g / k = (4.4 + 13.2 + 8.8) x 9.8 / 308 = 0.84 m = 84 cm

Hence, x' = 84 cm