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Chapter 6: Problem 28

A 0.60 -kg basketball is dropped out of a window that is \(6.1 \mathrm{~m}\) above the ground. The ball is caught by a person whose hands are \(1.5 \mathrm{~m}\) above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change \(\left(\mathrm{PE}_{\mathrm{f}} \mathrm{PE}_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?

Short Answer

Expert verified
(a) 27.05 J; (b) 35.87 J; (c) 8.82 J; (d) Work done equals change in potential energy.

Step by step solution

01

Calculate Work Done by Weight

Work done by weight (gravitational force) is determined by the change in gravitational potential energy as the basketball falls. The formula is \( W = m \cdot g \cdot (h_i - h_f) \), where \( m = 0.60 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( h_i = 6.1 \, \text{m} \), and \( h_f = 1.5 \, \text{m} \). Calculate \( W = 0.60\, \text{kg} \times 9.8\, \text{m/s}^2 \times (6.1\, \text{m} - 1.5\, \text{m}) = 0.60 \times 9.8 \times 4.6 = 27.048 \, \text{J} \).
02

Calculate Gravitational Potential Energy When Released

Gravitational potential energy \( (PE) \) relative to the ground is calculated using \( \text{PE} = m \cdot g \cdot h \). At release, \( h = 6.1 \, \text{m} \). Therefore, \( \text{PE}_{0} = 0.60 \times 9.8 \times 6.1 = 35.868 \, \text{J} \).
03

Calculate Gravitational Potential Energy When Caught

When the ball is caught at height \( 1.5 \, \text{m} \), the gravitational potential energy is \( \text{PE}_{f} = m \cdot g \cdot h = 0.60 \times 9.8 \times 1.5 = 8.82 \, \text{J} \).
04

Relate Change in Potential Energy to Work Done

The change in gravitational potential energy \((\Delta \text{PE})\) is defined as \( \text{PE}_{f} - \text{PE}_{0} \). Calculate this as \( 8.82\, \text{J} - 35.868\, \text{J} = -27.048\, \text{J} \). Notice the work done by the weight exactly equals this change, i.e., \( W = -\Delta \text{PE} \). This confirms that the work done by the weight equals the decrease in gravitational potential energy.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (PE) is a type of energy an object has due to its position above the ground. It depends on the height of the object and its mass.
When a basketball is held at a height, it has potential energy because of its ability to fall. This energy is called gravitational potential energy and can be calculated using the formula:
  • \(PE = m \cdot g \cdot h\)
where:
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)),
  • \(h\) is the height of the object above the ground.
When the basketball is released from a height of 6.1 m, the gravitational potential energy, compared to the ground level, is calculated as \(35.868 \, \text{J}\). This energy indicates the potential energy the basketball has to be converted once it starts falling.
Work Done by Gravity
When an object like the basketball falls, the gravitational force does work on it. This work is responsible for converting gravitational potential energy into kinetic energy as it descends.
The work done by gravity can be calculated using the formula:
  • \(W = m \cdot g \cdot (h_i - h_f)\)
Here,
  • \(h_i\) is the initial height,
  • \(h_f\) is the final height.
When the basketball falls from 6.1 m to 1.5 m,
  • the work done by the weight equals \(27.048 \, \text{J}\) in magnitude.
This means that 27.048 J of energy is transferred from gravitational potential to kinetic energy during the basketball's fall.
Conservation of Energy in Physics
One of the fundamental principles of physics is the conservation of energy. This principle states that the total energy in a closed system remains constant, though it may change forms. In our basketball example, the initial gravitational potential energy transforms into other forms as the ball falls.
  • The potential energy at the top is converted into kinetic energy as it falls.
The conservation of energy assures us that:
  • The work done by gravity on the ball equals the amount of potential energy lost.
  • As calculated, the change in gravitational potential energy (from release to being caught) aligns with the work done.
This illustrates perfectly how energy is conserved: potential energy is not lost but rather changes form, aligning with the work-energy principle that the work done equals the change in energy format. Thus, all the potential energy initially possessed by the basketball is accounted for as kinetic energy when it descends.

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Most popular questions from this chapter

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.

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by Lance Armstrong \((m=75.0)\) is \(6.50 \mathrm{~W}\) per kilogram of his body mass. (a) How much work does he do during a \(135-\mathrm{km}\) race in which his average speed is \(12.0 \mathrm{~m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

A shot-putter puts a shot (weight \(=71.1 \mathrm{~N}\) ) that leaves his hand at a distance of \(1.52 \mathrm{~m}\) above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of \(2.13 \mathrm{~m}\) above the ground. Include the correct sign for the work. (b) Determine the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy of the shot

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. (a) What type of energy is changing? (b) Is the work being done by the net external force acting on the skier positive, zero, or negative? Why? (c) How is this work related to the change in the energy of the skier? A 70.3 -kg water-skier has an initial speed of \(6.10 \mathrm{~m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{~m} / \mathrm{s}\). Determine the work done by the net external force acting on the skier.

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.
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