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A 0.60-kg basketball is dropped out of a window that is 6.1 m above the ground. The ball is caught by a person whose hands are 1.5 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 \(\left(PE_{\mathrm{f}}-PE_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?

Short Answer

Expert verified
The work done by gravity is 27.216 J. PE when released is 35.874 J; when caught is 8.82 J. The work done is equal to the negative change in gravitational potential energy.

Step by step solution

01

Understand the Problem

A basketball is dropped from 6.1 m and caught at 1.5 m. We need to calculate work done by gravity, and gravitational potential energy at different heights.
02

Identify Fundamental Formulas

The work done by gravity is given by the formula: \( W = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (\( 9.81\, m/s^2 \)), and \( h \) is the height difference. The gravitational potential energy (PE) is also calculated using \( PE = mgh \).
03

Calculate Work Done by Gravity

Work done by the ball's weight is calculated by finding the difference in potential energy from the release point to the catch point: \( W = mg(h_1 - h_2) = 0.60 \times 9.81 \times (6.1 - 1.5) \).Calculate this to find \( W \).
04

Calculate Initial Potential Energy

Calculate the gravitational potential energy when the ball is released:\( PE_0 = mgh = 0.60 \times 9.81 \times 6.1 \).
05

Calculate Final Potential Energy

Calculate the gravitational potential energy when the ball is caught:\( PE_f = mgh = 0.60 \times 9.81 \times 1.5 \).
06

Relating Work and Change in Potential Energy

The change in gravitational potential energy \( \Delta PE = PE_f - PE_0 \) is equal to the work done by the gravitational force (weight) on the basketball. So, \( W = \Delta PE \).

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Key Concepts

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

Work Done by Gravity
Work done by gravity is essentially the energy transferred to an object by the force of gravity as it moves between two points. To put it simply, it measures how much force gravity exerts on the object as it changes height. Gravity does the work whenever an object moves vertically along the direction of the force of gravity.

To calculate the work done by gravity, we use the formula:
  • \( W = mgh \)
where:
  • \( W \) is the work done by gravity
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity (approximately \( 9.81\, m/s^2 \) on Earth)
  • \( h \) is the height change the object undergoes
In the context of our basketball, as it falls from 6.1 m to 1.5 m above ground, gravity exerts a force throughout this height change. The work done by gravity converts potential energy at the starting height to kinetic energy and back to potential energy at the ending height. The formula showcases gravity’s role in energy transformation.
Energy Conservation
The principle of energy conservation is fundamental in physics, emphasizing that energy cannot be created or destroyed. It can only change forms. This is seen in the transition between potential and kinetic energy.

In the basketball example:
  • The basketball possesses potential energy due to its height.
  • As it falls, this potential energy converts into kinetic energy (energy of motion).
  • When the ball is caught, it again holds potential energy, but less than initially because it's closer to the ground.
The energy remains constant throughout, only its form changes. When the basketball drops:
  • Initial energy (when released) = gravitational potential energy
  • Midway during the fall = mix of potential and kinetic energy
  • Just before being caught = kinetic energy
Thus, the law of conservation of energy at play shows us how gravitational forces seamlessly exchange energy forms without losing any total energy in the isolated system of the ball and Earth.
Potential Energy Calculation
Calculating gravitational potential energy is like determining the stored energy an object possesses due to its position in a gravitational field. It’s energy held thanks to height above a reference point, typically the ground.

The formula used is:
  • \( PE = mgh \)
where:
  • \( PE \) stands for potential energy
  • \( m \) is the object's mass
  • \( g \) is the acceleration from gravity
  • \( h \) measures how high the object is from the reference point
Applying it to the basketball scenario:
  • When released from 6.1m, potential energy is calculated using \( PE_0 = 0.60 \times 9.81 \times 6.1 \).
  • When caught at 1.5m, it transforms to \( PE_f = 0.60 \times 9.81 \times 1.5 \).
This calculation reinforces the understanding of how height impacts stored gravitational energy. As the basketball falls to a lower height, its potential energy decreases, validating the role height plays in the energy retention of a lifting or falling mass.

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Most popular questions from this chapter

A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck skid before coming to a stop?

A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 24 N. Starting from rest, the sled attains a speed of 2.0 m/s in 8.0 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.500 s after leaving the slide. Ignoring friction and air resistance, find the height H in the drawing.

A person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is 94.0 N. (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down an head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped 126 m in elevation from top to bottom. (a) In the absence of non conservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed at the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with a speed of 40.5 m/s (about 91 mi/h). How much work was done on him and his sled (assuming a total mass of 118 kg) by non conservative forces during this heat?

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