A 54.0 kg woman steps off of a diving platform with a height of 12.0 m and drops straight down into the water. If she reaches a depth of 4.9 m, what is the average magnitude of the resistance force F average exerted on her by the water
The Correct Answer and Explanation is:
To solve this problem, we need to determine the average magnitude of the resistance force exerted by the water on the woman. We can apply the work-energy principle, which relates the change in kinetic energy to the work done by forces on an object. Here’s the step-by-step process:
Step 1: Determine the woman’s velocity just before hitting the water.
The woman falls freely from a height of 12.0 m. Using the equation for gravitational potential energy (PE), we can calculate her velocity just before hitting the water, assuming air resistance is negligible:
[
PE = mgh
]
Where:
- (m = 54.0 \, \text{kg}) is her mass,
- (g = 9.8 \, \text{m/s}^2) is the acceleration due to gravity,
- (h = 12.0 \, \text{m}) is the height of the platform.
The gravitational potential energy at the top of the dive is converted into kinetic energy (KE) at the moment she reaches the water’s surface:
[
KE = PE = mgh
]
Since kinetic energy is also expressed as:
[
KE = \frac{1}{2}mv^2
]
We can equate the two expressions for kinetic energy:
[
mgh = \frac{1}{2}mv^2
]
Solving for (v), the velocity just before hitting the water:
[
v = \sqrt{2gh} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 12.0 \, \text{m}} = \sqrt{235.2} \approx 15.3 \, \text{m/s}
]
Step 2: Calculate the woman’s velocity at the moment she reaches the maximum depth.
When the woman reaches a depth of 4.9 m, she has been decelerated by the water. At this point, her velocity is reduced to zero, as she comes to rest at the maximum depth.
Step 3: Apply the work-energy principle.
The work done by the water’s resistance force (F_{\text{average}}) is equal to the change in kinetic energy as she decelerates from her velocity just before impact to zero. The work done by the resistance force is:
[
W = F_{\text{average}} \times d
]
Where:
- (d = 4.9 \, \text{m}) is the distance the woman travels in the water.
The change in kinetic energy is:
[
\Delta KE = \frac{1}{2}mv^2 – 0 = \frac{1}{2} \times 54.0 \, \text{kg} \times (15.3 \, \text{m/s})^2 = 54.0 \times 117.0 \, \text{J} = 6320.0 \, \text{J}
]
The work done by the resistance force is equal to the change in kinetic energy:
[
F_{\text{average}} \times 4.9 \, \text{m} = 6320.0 \, \text{J}
]
Solving for (F_{\text{average}}):
[
F_{\text{average}} = \frac{6320.0 \, \text{J}}{4.9 \, \text{m}} \approx 1280.0 \, \text{N}
]
Final Answer:
The average magnitude of the resistance force exerted on the woman by the water is approximately 1280.0 N.
Explanation:
The solution uses principles of energy conservation and the work-energy theorem. Initially, the woman’s potential energy is converted to kinetic energy as she falls, and when she enters the water, the resistance force from the water does negative work to decelerate her. The magnitude of this force can be calculated using the change in kinetic energy and the distance she moves within the water. The resistance force is large because it is responsible for stopping the woman’s downward motion over a relatively short distance.