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Chapter 5: Problem 119

Suppose an Olympic diver who weighs \(52.0 \mathrm{~kg}\) executes a straight dive from a 10 -m platform. At the apex of the dive, the diver is $10.8 \mathrm{~m}$ above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in $\mathrm{m} / \mathrm{s}$, will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

Short Answer

Expert verified
The potential energy of the diver at the apex of the dive is approximately 5529.216 Joules. The diver will enter the water at a speed of approximately 14.27 m/s. Yes, the diver does work on entering the water, as they transfer energy to the water through the force they apply while entering.

Step by step solution

01

Part (a): Calculate the potential energy at the apex of the dive

To calculate the potential energy (PE) of the diver at the peak of the dive, we can use the potential energy formula: \(PE = mgh\) Where: \(m = 52.0\text{ kg}\) (mass of the diver) \(g = 9.81 \frac{\text{m}}{\text{s}^2}\) (acceleration due to gravity) \(h = 10.8 \text{ m}\) (height above the water surface) Plugging in the values we have: \(PE = (52.0\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})(10.8\text{ m})\)
02

Part (a): Compute the potential energy

Now, we can compute the potential energy of the diver at the peak of the dive: \(PE = 5529.216\text{ J}\) So, the potential energy of the diver at the apex of the dive, relative to the surface of the water, is approximately 5529.216 Joules.
03

Part (b): Use conservation of mechanical energy

We can assume the mechanical energy of the diver is conserved during the dive. This means that the potential energy at the highest point is all converted into kinetic energy (KE) just before entering the water. Thus, \(PE = KE\) We can use the kinetic energy formula to find the speed of the diver at the water surface: \(KE = \frac{1}{2}mv^2\) Where: \(v\) is the speed of the diver, and the other variables have been defined earlier.
04

Part (b): Calculate the speed at the surface of the water

Now, we can solve for the speed (\(v\)) of the diver just before entering the water: \(5529.216\text{ J} = \frac{1}{2}(52.0\text{ kg})v^2\) Now we find v: \(v = \sqrt{\frac{2 \times 5529.216\text{ J}}{52.0\text{ kg}}}\)
05

Part (b): Compute the dive speed

Finally, we can compute the speed of the diver as they enter the water: \(v \approx 14.27\frac{\text{m}}{\text{s}}\) So, the diver will enter the water at a speed of approximately 14.27 m/s.
06

Part (c): Does the diver do work on entering the water?

When the diver enters the water, the water applies a force on the diver to slow them down. This force is in the opposite direction of the diver's motion, so it does negative work on the diver. In turn, the diver applies force on the water as they enter it (Newton's third law), which means the diver does work on the water. This work done by the diver on the water results in the transfer of energy from the diver to the water (as some of the diver's kinetic energy is lost to heat, sound, and water displacement). So, yes, the diver does work on entering the water.

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