/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Refer to Interactive Solution \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Refer to Interactive Solution \(6.51\) for a review of the approach taken in problems such as this one. A \(67.0\) -kg person jumps from rest off a \(3.00\) -m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest \(1.10 \mathrm{~m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

Short Answer

Expert verified
The average force exerted by the water on the diver is approximately 1789.64 N.

Step by step solution

01

Understand Potential Energy on the Tower

When the person is on top of the tower, they have gravitational potential energy given by the formula \( PE = mgh \), where \( m = 67.0 \) kg is the person's mass, \( g = 9.8 \) m/s² is the acceleration due to gravity, and \( h = 3.00 \) m is the height of the tower. Calculate the potential energy: \( PE = 67.0 \times 9.8 \times 3.00 \).
02

Calculate Potential Energy

Substitute the values into the potential energy formula from Step 1: \( PE = 67.0 \times 9.8 \times 3.00 = 1968.6 \) J (joules). This is the energy the person has just before entering the water.
03

Determine Kinetic Energy Upon Entering the Water

As the diver falls, the potential energy is converted to kinetic energy. Right before they enter the water, their kinetic energy equals their initial potential energy (1968.6 J).
04

Calculate Work Done by Water

When the diver enters the water and slows to a stop, the water does work on the diver, which equals the change in kinetic energy. The work done by the water (\( W \)) is equal to the initial kinetic energy (1968.6 J) since the final kinetic energy is zero.
05

Apply Work-Energy Principle

The work done by the water is also calculated by the equation \( W = F_{average} \times d \), where \( F_{average} \) is the average force exerted by the water, and \( d = 1.10 \) m is the distance over which water exerts force. Rearrange this formula to find the average force: \( F_{average} = \frac{W}{d} \).
06

Calculate Average Force Exerted by the Water

Substitute the values from the previous steps into the formula from Step 5: \( F_{average} = \frac{1968.6}{1.10} = 1789.64 \) N. This is the magnitude of the average force the water exerts on the diver.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy, often abbreviated as PE, is the stored energy in an object due to its position or configuration. For a diver on a high tower, this energy is in the form of gravitational potential energy. It depends on three factors:
  • Mass (\( m \)): The amount of matter in the diver's body.
  • Gravitational acceleration (\( g \)): A constant \( 9.8 \, \text{m/s}^2 \), which is the rate at which gravity accelerates objects downward.
  • Height (\( h \)): The vertical distance above the water.
The formula used to calculate potential energy is \( PE = mgh \).
For the diver, this calculation turns into \( PE = 67.0 \times 9.8 \times 3.00 \), giving us \( 1968.6 \) joules just before entering the water.
The concept of potential energy helps us understand the energy transformation that occurs as the diver falls. As she descends, this stored energy begins to change into kinetic energy.
Kinetic Energy
Kinetic energy describes the energy an object has due to its motion. Once the diver leaps from the tower, her potential energy is converted into kinetic energy as she accelerates towards the water.
Just before hitting the water, all her initial potential energy is transformed into kinetic energy. This is because energy conservation implies that no energy is lost or gained, merely transformed from one form to another.
We use the same value calculated for potential energy to represent her kinetic energy right before impact: \( KE = 1968.6 \, \text{J} \).
This kinetic energy explains the power and speed with which the diver enters the water, illustrating a fundamental principle: what goes up, must come down, and the energy does not disappear during this process.
Nonconservative Forces
Nonconservative forces are forces that depend on the path taken and involve energy dissipation, such as friction or air resistance. In the case of the diver, the water's resistance as she penetrates the surface acts as a nonconservative force.
These forces take kinetic energy out of the diver, causing her to decelerate until she comes to a stop underwater. This energy is transformed rather than conserved, and cannot be described by only the initial and final states of motion.
The water force functioned here in dissipating the diver's kinetic energy into heat and sound as she slows from her initial speed to zero.
  • Unlike gravity, which is conservative, water resistance is nonconservative because the energy is not stored for reuse.
  • The Work-Energy Principle relates here, showing that the work done by a nonconservative force equals the change in kinetic energy, which was zero when the diver stopped.
Average Force
The average force is a concept used to describe a constant force that yields the same energy change as the actual variable force over a distance. We determine this force during instances like a diver slowing in water due to nonconservative forces.
Following the Work-Energy Principle, the work done by the water \( (W) \) is equivalent to the diver's kinetic energy just before hitting the water: \( 1968.6 \, \text{J} \).
  • Distance (\( d \)) over which this force acts is \( 1.10 \, \text{m} \)
  • The formula for calculating average force is \( F_{average} = \frac{W}{d} \)
  • Substituting the values: \( F_{average} = \frac{1968.6}{1.10} = 1789.64 \, \text{N} \)
This average force provides insight into the force exerted by the water against the diver, acting over the distance taken to bring her motion to rest.

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Most popular questions from this chapter

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. (a) What type (s) of energy is (are) changing? Is each type increasing or decreasing? Why? (b) How is (are) the type(s) of energy that is (are) changing related to the work done by the lifting force? (c) If you want to determine the average power generated by the lifting force, what other variable besides the work must be known? An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.5 \mathrm{~s}\). During this time it climbs to a height of \(8.2 \mathrm{~m}\). What is the average power generated by the lifting force?

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of \(8.90 \mathrm{~m} / \mathrm{s},\) and the second lands at a speed of \(9.00 \mathrm{~m} / \mathrm{s} .\) The first vaulter clears the bar at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. (a) Does the kinetic frictional force do positive, negative, or zero work? Provide a reason for your answer. (b) Does the total mechanical energy of the student increase, decrease, or remain the same as she descends the slide? Why? (c) If the kinetic frictional force does work, how is this work related to the change in the total mechanical energy of the student? The student has a mass of \(83.0 \mathrm{~kg}\) and the height of the water slide is \(11.8 \mathrm{~m}\). If the kinetic frictional force does \(-6.50 \times 10^{3} \mathrm{~J}\) of work, how fast is the student going at the bottom of the slide?

During a tug-of-war, team A pulls on team B by applying a force of \(1100 \mathrm{~N}\) to the rope between them. How much work does team A do if they pull team B toward them a distance of \(2.0 \mathrm{~m} ?\)

Refer to Concept Simulation 6.1 at for a review of the concepts with which this problem deals. A \(0.075-\mathrm{kg}\) arrow is fired horizontally. The bowstring exerts an average force of \(65 \mathrm{~N}\) on the arrow over a distance of \(0.90 \mathrm{~m}\). With what speed does the arrow leave the bow?

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