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Chapter 6: Problem 83

A basketball of mass 0.60 kg is dropped from rest from a height of 1.05 m. It rebounds to a height of 0.57 m. (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of 1.05 m by exerting a constant downward force on it for a distance of 0.080 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) 2.824 J lost. (b) Force required is approximately 35.30 N.

Step by step solution

01

Calculate Initial Potential Energy

The initial potential energy (PE) of the basketball can be calculated using the formula \(PE = mgh\), where \(m\) is the mass of the basketball, \(g\) is the acceleration due to gravity (approximately 9.8 m/s虏), and \(h\) is the initial height. Given \(m = 0.60\, \text{kg}\) and \(h = 1.05\, \text{m}\), the initial potential energy is:\[ PE_1 = 0.60 \times 9.8 \times 1.05 \approx 6.174 \text{ J}. \]
02

Calculate Final Potential Energy

After rebounding, the potential energy when the basketball reaches its maximum height of 0.57 m is calculated similarly: \[ PE_2 = 0.60 \times 9.8 \times 0.57 \approx 3.3504 \text{ J}. \]
03

Determine Mechanical Energy Lost

The mechanical energy lost during the collision is the difference between the initial and final potential energies: \[ \Delta E = PE_1 - PE_2 = 6.174 \text{ J} - 3.3504 \text{ J} = 2.8236 \text{ J}. \]
04

Calculate Work Done by the Player's Force

The player needs to compensate for the lost mechanical energy by doing an equivalent amount of work on the ball. The work done by a constant force over a distance is given by \(W = F \times d\), where \(F\) is the magnitude of the force and \(d = 0.080\, \text{m}\) is the distance over which the force is applied. Therefore, setting \(W = \Delta E\), we have:\[ F \times 0.080 = 2.8236 \text{ J}. \]
05

Solve for the Force

To find the force, divide the energy difference by the distance:\[ F = \frac{2.8236}{0.080} \approx 35.295 \text{ N}. \]

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Key Concepts

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

Mechanical Energy Loss
When a basketball is dropped and hits the ground, some of its mechanical energy is lost in the process. This loss occurs primarily due to the conversion of mechanical energy into other forms, such as sound and thermal energy. In our exercise, when the basketball falls from 1.05 m and bounces back to only 0.57 m, we calculate the mechanical energy loss by finding the difference in its potential energy before and after the bounce.
Considering the initial and final potential energy values from our problem, we found:
  • Initial potential energy: 6.174 J
  • Final potential energy: 3.3504 J
  • Mechanical energy lost: 2.8236 J
This means that 2.8236 Joules of energy were not retained in the form of potential energy after the basketball's collision with the floor.
Potential Energy
Potential energy is the stored energy in an object due to its position relative to a reference point. In the context of our exercise, potential energy is determined by the height at which the basketball is positioned above the ground.
Potential energy is calculated with the formula: \[ PE = mgh \]where
  • \(m\) is the mass of the object (0.60 kg for our basketball)
  • \(g\) is the acceleration due to gravity (approximately 9.8 m/s虏)
  • \(h\) is the height above the reference point (initially 1.05 m and 0.57 m after rebounding)
Potential energy plays a crucial role in energy conservation exercises as it is a direct measure of how much energy an object can potentially release under gravity's influence.
Work-Energy Principle
The work-energy principle links the concepts of work and energy by stating that the work done on an object is equal to the change in its kinetic energy. In mechanics, this principle helps us understand how forces affect an object's motion and energy state.
In the dribbling scenario from the exercise, the player must exert work equal to the energy lost during the basketball's bounce to make it return to its original height. This work compensates for lost potential energy by adding kinetic energy back into the system when the ball hits the ground.
  • The player applies a force over a distance (0.080 m in our calculation)
  • The work done \(W\) is given by the equation: \[ W = F \times d \]
  • The magnitude of the force \(F\) determined from the energy lost is approximately 35.295 N
By balancing work and energy, the player ensures the ball returns to its original height, overcoming mechanical energy losses.
Collision Mechanics
In collision mechanics, we study interactions where objects come into contact and exert forces upon one another. A basketball colliding with the floor showcases a simple example of such interactions where energy transformations play a central role.
During a bounce, not all of the basketball's energy is preserved as potential energy; some is lost in forms like sound or deformation. The collision mechanics in our problem demonstrate this by showing a decrease in potential energy from 6.174 J to 3.3504 J.
  • Collisions often exhibit energy loss that we calculate through changes in potential energy.
  • Understanding these energy changes can inform us about the material properties of the basketball and the floor.
  • Players need to understand these mechanics to maintain force control during dribbling for effective play.
These concepts are not just theoretical but have practical implications in sports and other fields where controlling the outcome of collisions is crucial.

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