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The first step here is to find the spring constant. Use Hooke's law and the information given about the 3 kg mass:

F=kx

F=w3kg-mass=m3kg-massg=(3 kg)(9.8 m/s2)

(3 kg)(9.8 m/s2)=(.40 m)k

k=[(3 kg)(9.8 m/s2)/(.40 m)]

Now plug that in (I don't have a calculator handy, and some of the units and numbers there will cancel, so I didn't bother to calculate it out) to Hooke's law for the 5 kg mass:

F=kx

F=w5kg-mass=m3kg-massg=(5 kg)(9.8 m/s2)

(5 kg)(9.8 m/s2)=kx

(5 kg)(9.8 m/s2)=[(3 kg)(9.8 m/s2)/(.40 m)]x

x=2/3 m

The 5 kg mass would stretch the spring two thirds of a meter.

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