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**Problem 1:**

When a 4 kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5 cm. If the 4 kg mass is removed,(a) how far will the spring stretch if a 1.5 kg mass is hung on it, and

(b) how much work must an external agent do to stretch the same spring 4 cm from its unstretched position?

- Solution:(a) We find the spring constant of the spring from the given data.

F = -kx.

F = -mg = -(4 kg)(9.8 m/s^{2) }= -39.2 N.

k = F/x = (39.2 N)/(0.025 m) = 1568 N/m.

Now we use x = -F/k to find the displacement of a 1.5 kg mass.

F = -(1.5kg)(9.8m.s^{2}) = -14.7 N.

x = (14.7 N)/(1568 N/m) = 0.009375 m = 0.975 cm.

(b) W = (1/2)kx^{2 }= (1/2)(1568 N/m)(0.04 m)^{2 }= 1.2544 Nm = 1.2544 J

#### Problem 2:

If it takes 4J of work to stretch a Hooke's law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.- Solution:

The work done in stretching or compressing a spring is proportional to the square of the displacement. If we double the displacement, we do 4 times as much work. It takes 16 J to stretch the spring 20 cm from its unstretched length, so it takes 12 J to stretch it from 10 cm to 20 cm.Formally:

W = (1/2)kx^{2}. Given W and x we find k.

4 J = (1/2)k(0.1 m)^{2}; k =(8 J)/(0.1 m)^{2 }= 800 N/m.

Now x = 0.2 m. W = (1/2)(800 N/m)(0.2 m)^{2 }= 16 J.

Extra work: 16 J - 4 J = 12 J.