Problem 21P

Estimate the density of the water 6.0 km deep in the sea. (See Table 9–1 and Section 9–5 regarding bulk modulus.) By what fraction does it differ from the density at the surface?

TABLE 9–1 Elastic Moduli

ANSWER:STEP 1:-Every ideal spring obeys Hooke’s law. The law tells about the relation between thedisplacement of the spring with the restoring force.The force is, F = Kx-------------------(1)The potential energy whose derivative will give such force is, P.E = 1/2 Kx -----------------(2)Here K is the spring constant and x is the displacement of the spring.STEP 2:-The sprinters stretch 33 mm which is equal to 0.001 × 41 = 0.041 metersThe spring constant given for them is, 33 N/mmAs we know, 1mm = 0.001 m 33 N/mm = 33/0.001 = 33000 N/mSo, K=33000 N/m.The potential energy stored is, P.E = 1/2 Kx = 1/2 × 33000 × 0.041 = 27.73 jules.STEP 3:-The nonathletes stretch 41 mm which is equal to 0.001 × 33 = 0.033 metersThe spring constant given for them is, 33 N/mm\nAs we know, 1mm = 0.001 m 33 N/mm = 33/0.001 = 33000 N/mSo, K=33000 N/m.The potential energy stored is, P.E = 1/2 Kx = 1/2 × 33000 × 0.033 = 17.96 = 18jules (approximately).STEP 4:-The difference between the potential energies is, P.E = 27.73 18 = 9.73 jules.CONCLUSION:The difference between maximum stored energies between the sprinter and nonathletes is9.73 jules.