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A \(75 \mathrm{~g}\) Frisbee is thrown from a point \(1.1 \mathrm{~m}\) above the ground with a speed of \(12 \mathrm{~m} / \mathrm{s}\). When it has reached a height of \(2.1 \mathrm{~m},\) its speed is \(10.5 \mathrm{~m} / \mathrm{s} .\) What was the reduction in \(E_{\mathrm{mec}}\) of the Frisbee-Earth system because of air drag?

Short Answer

Expert verified
The reduction in mechanical energy due to air drag is \(0.517 \, \text{J}\).

Step by step solution

01

Analyze Given Values

The problem provides the following values: initial speed \( v_i = 12 \, \text{m/s} \), initial height \( h_i = 1.1 \, \text{m} \), final speed \( v_f = 10.5 \, \text{m/s} \), final height \( h_f = 2.1 \, \text{m} \), and the mass of the Frisbee \( m = 75 \, \text{g} = 0.075 \, \text{kg} \).
02

Calculate Initial Mechanical Energy

The mechanical energy consists of kinetic and gravitational potential energy. Using the formulas \( KE = \frac{1}{2}mv^2 \) and \( PE = mgh \), calculate the initial mechanical energy: \[ E_{\text{mec,i}} = KE_i + PE_i = \frac{1}{2} \times 0.075 \, \text{kg} \times (12 \, \text{m/s})^2 + 0.075 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.1 \, \text{m} \] Calculate the initial kinetic energy: \[ KE_i = \frac{1}{2} \times 0.075 \times 144 = 5.4 \, \text{J} \] Calculate the initial potential energy: \[ PE_i = 0.075 \times 9.81 \times 1.1 = 0.809 \, \text{J} \] \[ E_{\text{mec,i}} = 5.4 + 0.809 = 6.209 \, \text{J} \]
03

Calculate Final Mechanical Energy

Similarly, calculate the final mechanical energy: \[ E_{\text{mec,f}} = KE_f + PE_f = \frac{1}{2} \times 0.075 \, \text{kg} \times (10.5 \, \text{m/s})^2 + 0.075 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 2.1 \, \text{m} \] Calculate the final kinetic energy: \[ KE_f = \frac{1}{2} \times 0.075 \times 110.25 = 4.144 \, \text{J} \] Calculate the final potential energy: \[ PE_f = 0.075 \times 9.81 \times 2.1 = 1.548 \, \text{J} \] \[ E_{\text{mec,f}} = 4.144 + 1.548 = 5.692 \, \text{J} \]
04

Calculate Mechanical Energy Reduction Due to Air Drag

The reduction in mechanical energy due to air drag is the difference between the initial and final mechanical energies. \[ \Delta E_{\text{mec}} = E_{\text{mec,i}} - E_{\text{mec,f}} = 6.209 \, \text{J} - 5.692 \, \text{J} \] \[ \Delta E_{\text{mec}} = 0.517 \, \text{J} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It can be thought of as the energy needed to move an object from rest to its current speed. In the context of the Frisbee problem, we calculate kinetic energy using the formula:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity. Kinetic energy increases with the square of the velocity, meaning if the speed of an object doubles, its kinetic energy increases by a factor of four. This relationship is crucial in understanding how changes in motion can affect energy conservation.

The initial kinetic energy of the Frisbee was calculated from its initial speed of 12 m/s. As the Frisbee ascended and its speed decreased to 10.5 m/s, its kinetic energy was reduced. This transformation illustrates the conversion of kinetic energy into potential energy and the overall interplay of forces like air resistance impacting mechanical energy.
Potential Energy
Potential energy is the energy stored in an object due to its position or arrangement. For objects near the Earth's surface, gravitational potential energy is a common form and is mostly influenced by height relative to a reference point.

It is calculated using the formula:
  • \( PE = mgh \)
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and \( h \) is the height above the reference point.

In the exercise, as the Frisbee was thrown upwards, its height increased from 1.1 m to 2.1 m. Consequently, its potential energy increased, while kinetic energy decreased. This transformation highlights the conservation of mechanical energy when no external work is done, or energy is dissipated due to non-conservative forces like air drag. Understanding this process helps in identifying energy shifts and energy losses in real-world scenarios.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system. It represents the total energy available for doing work in mechanical systems. In an ideal scenario, with no external forces like friction or air resistance, the mechanical energy of a system remains constant. However, real-world conditions often involve energy dissipation which affects this constancy.

In our problem, the initial mechanical energy of the Frisbee was a combination of its initial kinetic energy and potential energy. This was calculated as:
  • \( E_{\text{mec,i}} = KE_i + PE_i = 5.4 \, \text{J} + 0.809 \, \text{J} = 6.209 \, \text{J} \)

Similarly, the final mechanical energy was:
  • \( E_{\text{mec,f}} = KE_f + PE_f = 4.144 \, \text{J} + 1.548 \, \text{J} = 5.692 \, \text{J} \)

The reduction in mechanical energy, calculated as 0.517 J, indicated the energy lost due to air drag. This real-world loss illustrates how external forces impact the conservation laws, providing insight into energy transformations and conservation in everyday physics problems.

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Most popular questions from this chapter

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20 .\) What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

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