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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 \(1.5 \mathrm{m}\) 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
The work done by gravity is the difference in potential energy from 6.1 m to 1.5 m. The change in potential energy is equal to negative the work done by gravity.

Step by step solution

01

Understanding Work Done by Gravity

The work done by gravity on the basketball can be calculated using the formula for work: \[ W = F \cdot d \cdot \cos(\theta) \]Since gravity does the work and acts in the direction of motion, \(\theta = 0\) degrees and \(\cos(0) = 1\), so:\[ W = F \cdot d = m \cdot g \cdot d \]Here, \(m = 0.60\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity, and \(d = 6.1\,\text{m} - 1.5\,\text{m} = 4.6\,\text{m}\).
02

Calculating Work Done

Using the values identified:\[ W = 0.60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 4.6 \, \text{m} \]Calculate the result to find the work done by gravity.
03

Calculating Initial Potential Energy

Gravitational potential energy (PE) is given by the formula:\[ \text{PE} = m \cdot g \cdot h \]For the ball at its release height \(h = 6.1\, \text{m}\):\[ \text{PE}_{\text{initial}} = 0.60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 6.1 \, \text{m} \]
04

Calculating Final Potential Energy

The potential energy when the ball is caught with height \(h = 1.5\, \text{m}\) is:\[ \text{PE}_{\text{final}} = 0.60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.5 \, \text{m} \]
05

Assessing Change in Potential Energy

The change in potential energy of the ball is:\[ \Delta \text{PE} = \text{PE}_{\text{final}} - \text{PE}_{\text{initial}} \]
06

Relating Work and Potential Energy

The work done by the weight of the ball is equal to the negative change in potential energy:\[ W = -(\Delta \text{PE}) \]This indicates that the energy lost due to work done by gravity is equal to the decrease in 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 (GPE) is a concept used to describe the energy that an object possesses due to its position in a gravitational field. It is an essential idea in physics as it helps us understand how energy is stored and transformed. The formula for gravitational potential energy is given by:
\[ \text{PE} = m \cdot g \cdot h \]
where:
  • \( m \) is the mass of the object (measured in kilograms)
  • \( g \) is the acceleration due to gravity \((9.8 \, \text{m/s}^2)\)
  • \( h \) is the height above the reference point (usually the ground) measured in meters.
The initial potential energy when the basketball is released from a height of 6.1 meters can be calculated by substituting these values into the formula. Similarly, when the basketball is caught at a height of 1.5 meters, the potential energy decreases because the height is reduced. These calculations illustrate how gravitational potential energy is directly proportional to the height at which the object is located.
Work Done by Gravity
Work done by gravity is a fundamental principle that describes how gravity affects the movement of an object. In simpler terms, it is the energy transferred from gravity to the object as it moves downward. This is calculated by using the work formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Since gravity acts in the direction of motion (downwards), \( \theta \) is 0 degrees, and \( \cos(0) = 1 \), simplifying the formula to:
\[ W = m \cdot g \cdot d \]This formula represents the product of the object's weight and the distance it moves in the direction of the force. In our scenario, the basketball moves from 6.1 meters to 1.5 meters. Therefore, the distance \( d \) is 4.6 meters.
The work done by gravity in this context is the gravitational force acting on the basketball causing it to move downward a certain distance. This work done is effectively the energy transfer that lowers the basketball's potential energy as it falls.
Energy Conservation
Energy conservation is a core principle of physics, emphasizing that energy cannot be created or destroyed, only transformed from one form to another. In the context of the basketball scenario, energy is transformed from gravitational potential energy to kinetic energy and vice versa.
When the basketball is initially dropped, it holds gravitational potential energy due to its position. As it falls, this energy is converted into kinetic energy (the energy of motion). By the time the basketball is caught, the potential energy is lower due to its reduced height, and kinetic energy is at a maximum.
The relationship between the work done by gravity and the change in gravitational potential energy is expressed as:
  • Work done by gravity = negative change in potential energy (\( W = -\Delta \text{PE} \))
This equation demonstrates the principle of energy conservation by showing that the decrease in potential energy corresponds exactly to the work done by gravity, confirming that energy within the system remains constant overall.

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Most popular questions from this chapter

A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{m} / \mathrm{s} .\) The building is \(31.0 \mathrm{m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

A semitrailer is coasting downhill along a mountain highway when its brakes fail. The driver pulls onto a runaway-truck ramp that is inclined at an angle of \(14.0^{\circ}\) above the horizontal. The semitrailer coasts to a stop after traveling \(154 \mathrm{m}\) along the ramp. What was the truck's initial speed? Neglect air resistance and friction.

A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{m}\). She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0-\mathrm{N}\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

At a carnival, you can try to ring a bell by striking a target with a \(9.00-\mathrm{kg}\) hammer. In response, a \(0.400-\mathrm{kg}\) metal piece is sent upward toward the bell, which is \(5.00 \mathrm{m}\) above. Suppose that \(25.0 \%\) of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

A cable lifts a 1200 -kg elevator at a constant velocity for a distance of \(35 \mathrm{m} .\) What is the work done by (a) the tension in the cable and (b) the elevator's weight?

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