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dancerina

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"A 63 kg diver is standing on a diving board waiting to commence her dive. The diving board has compressed 8.0 cm. Determine the elastic potential energy of the diver."

The solution given by the book is to consider the diving board as a spring and set the net force as zero, with the forces being that of the diver's weight and the spring's resistive force:

Fnet = mg - kx = 0

so k = (63 kg)(9.8 m/s^2)/(0.08 m) = 7717.5 kg/s^2

Then Ee = (1/2)kx^2 = (1/2)(7717.5 kg/s^2)(0.08m)^2 = 24.696 J

However, some students have said that the gravitational potential energy of the diver before they compress the board should be equal to the elastic potential energy that gets stored in the board; i.e.:

Eg = Ee, so mgh = Ee = (63 kg)(9.8 m/s^2)(0.08m) = 49.392 J (exactly double the previous answer).

They argue that the first method means 1/2 the Eg is lost somewhere (we're assuming this is an ideal situation). Can someone explain why this second method is incorrect?