The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

a. How much gravitational potential energy does the diver have?

b. How much kinetic energy does the diver have as he strikes the water?

c. How much work does the water do on the diver to stop his momentum?

d. If the diver stops in 2.5 m after entering the water, what net force is applied to the diver?

The gravitational potential energy of the diver is 6000 J.

The gravitational potential energy = mgh

m = mass of the diver = 60 kg

g = acceleration due to gravity = 10 ms-2

h = height =  10 m

gravitational potential energy = 60 kg ×  10 ms-2 × 10 m = 6000 J

Recall that the potential energy of the diver = kinetic energy does the diver have as he strikes the water. Hence, kinetic energy does the diver have as he strikes the water is 6000J.

The work done by the water= gravitational potential energy.

Hence, the  work done by the water on the diver to stop his momentum = 6000 J.

Recall that; work done = Force × distance

Force = work done/distance

Force = 6000J/2.5 m

Force = 2400 N

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a) Ep = 5886[J]; b) v = 14[m/s]; c)   W = 5886[J]; d) F = 1763.4[N]

Explanation:

a)

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

b)

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

c)

The work is equal to

W = 5886 [J]

d)

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

c) up

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