## Tuesday, 16 September 2014

### ELASTIC POTENTIAL ENERGY PROBLEMS ND SOLUTIONS

#### 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/s2= -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.s2= -14.7 N.
x = (14.7 N)/(1568 N/m) = 0.009375 m = 0.975 cm.
(b) W = (1/2)kx2 = (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)kx2.  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.