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Chapter 6: Problem 83

A basketball of mass \(0.60 \mathrm{kg}\) is dropped from rest from a height of \(1.05 \mathrm{m}\). It rebounds to a height of \(0.57 \mathrm{m}\). (a) How much mechanical energy was lost during the collision with the floor? (b) \(\mathrm{A}\) basketball player dribbles the ball from a height of \(1.05 \mathrm{m}\) by exerting a constant downward force on it for a distance of \(0.080 \mathrm{m}\). In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 m, what is the magnitude of the force?

Short Answer

Expert verified
(a) 2.83 J of energy was lost; (b) the force exerted by the player is 35.35 N.

Step by step solution

01

Calculate Initial Potential Energy

Calculate the initial potential energy when the basketball is at a height of 1.05 m:\[PE_{initial} = mgh_{initial} = 0.60 \, ext{kg} \times 9.81 \, \text{m/s}^2 \times 1.05 \, \text{m}\]
02

Calculate Final Potential Energy

Calculate the final potential energy after the basketball rebounds to a height of 0.57 m:\[PE_{final} = mgh_{final} = 0.60 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.57 \, \text{m}\]
03

Compute the Mechanical Energy Lost

Find the lost mechanical energy during the collision by subtracting the final potential energy from the initial potential energy:\[\Delta E = PE_{initial} - PE_{final}\]
04

Calculate Energy Required to Compensate Loss

The player compensates for the energy lost, so we calculate the energy difference needed to return the ball to the original height:\[E_{compensate} = PE_{initial} - PE_{final}\]
05

Determine Work Done by the Player

The player exerts a force over a distance of 0.080 m, and this work is equal to the energy compensated for:\[W = F \cdot d = E_{compensate}\]Solve for the force \(F\):\[F = \frac{E_{compensate}}{d} = \frac{PE_{initial} - PE_{final}}{0.080 \, \text{m}}\]
06

Perform Calculations for Part (a)

Substitute the given values to calculate the initial and final potential energy:1. \(PE_{initial} = 0.60 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.05 \, \text{m} = 6.183 \, \text{J}\)2. \(PE_{final} = 0.60 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.57 \, \text{m} = 3.3552 \, \text{J}\)3. \(\Delta E = 6.183 \, \text{J} - 3.3552 \, \text{J} = 2.8278 \, \text{J}\)
07

Perform Calculation for Part (b)

Use the energy loss to calculate the force exerted by the player:\[F = \frac{2.8278 \, \text{J}}{0.080 \, \text{m}} = 35.3475 \, \text{N}\]

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Key Concepts

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

Potential Energy
Potential energy is the energy possessed by an object due to its position relative to other objects. For a basketball at height, we calculate it using the formula:
  • \(PE = mgh\)
where:
  • \(m\) is mass,
  • \(g\) is gravitational acceleration (9.81 \(\text{m/s}^2\)),
  • \(h\) is height.

Initially, our basketball has a high potential energy when dropped from 1.05 m. This energy decreases as the ball falls and rebounds to a lower height (0.57 m), resulting in lower final potential energy. It's crucial to recognize how height plays a pivotal role in determining potential energy, as the difference in starting and rebounding height directly affects the energy change.
Mechanical Energy
Mechanical energy is the sum of potential and kinetic energy in an object. In our situation, we focus on the change in potential energy to understand the overall mechanical energy change.
Although energy should ideally be conserved in a perfect collision, some mechanical energy is lost. This loss happens due to factors like friction and deformation during the impact between the ball and the floor.
  • The initial mechanical energy of our system is when the ball is stationary at its initial height.
  • After rebounding, the mechanical energy is lower because the ball doesn't reach its original height again.
Understanding mechanical energy helps us figure out how much energy is transferred or lost in different processes, showcasing the foundational principle of the conservation of energy.
Energy Loss
Energy loss refers to the reduction in mechanical energy when the basketball hits the floor and doesn't bounce back to its original height.
  • Energy loss = Initial Potential Energy - Final Potential Energy
In our scenario, the energy lost is around 2.8278 J. This energy loss can occur due to:
  • Sound produced during the bounce,
  • Thermal energy from friction with the floor,
  • Internal damping resulting in deformation.
Recognizing where energy goes helps us appreciate how real-world applications deviate from ideal scenarios, such as frictionless surfaces.
Force Calculation
When a basketball player dribbles a ball, they apply a force to compensate for the energy lost during the bounce. To return the ball to the original height, this force makes up for the lost mechanical energy.
The formula used here is:
  • \(F = \frac{E_{compensate}}{d}\)
where:
  • \(F\) is the force,
  • \(E_{compensate}\) is the energy needed to return to the original height (2.8278 J),
  • \(d\) is the distance over which the force is applied (0.080 m).
By calculating it, we determine that a force of approximately 35.3475 N needs to be exerted. This calculation shows how forces interact with mechanical work to resolve energy losses in physical activities.

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Most popular questions from this chapter

Starting from rest, a \(1.9 \times 10^{-4}-\mathrm{kg}\) flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of \(0.38 \mathrm{N}\) on it. This force does \(+2.4 \times 10^{-4} \mathrm{J}\) of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{N}\). What is the magnitude of the pulling force exerted by his wife?

A projectile of mass 0.750 kg is shot straight up with an initial speed of \(18.0 \mathrm{m} / \mathrm{s}\) (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only \(11.8 \mathrm{m}\), determine the magnitude of the average force due to air resistance.

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of \(0.75 \mathrm{m}\). With the string hanging vertically, the object is given an initial velocity of \(2.0 \mathrm{m} / \mathrm{s}\) parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle \(\theta\) with its initial vertical orientation and then swings back downward. Find the angle \(\theta\)

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