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Chapter 8: Problem 47

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 approximately 0.93 J.

Step by step solution

01

Understanding the Problem

We're given a Frisbee with mass of 75 g (or 0.075 kg) thrown at a speed of 12 m/s from a height of 1.1 m. We need to find the reduction in mechanical energy \( E_{\text{mec}} \) when the Frisbee reaches a height of 2.1 m with a speed of 10.5 m/s. The reduction in energy is due to air drag.
02

Calculate Initial Mechanical Energy

The initial mechanical energy \( E_{\text{initial}} \) is the sum of initial kinetic energy \( K_i \) and initial potential energy \( U_i \): \[ K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.075 \times (12)^2 \] \[ U_i = mgh_i = 0.075 \times 9.8 \times 1.1 \] Calculate \( K_i \) and \( U_i \) to find \( E_{\text{initial}} = K_i + U_i \).
03

Calculate Final Mechanical Energy

The final mechanical energy \( E_{\text{final}} \) is the sum of final kinetic energy \( K_f \) and final potential energy \( U_f \): \[ K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.075 \times (10.5)^2 \] \[ U_f = mgh_f = 0.075 \times 9.8 \times 2.1 \] Calculate \( K_f \) and \( U_f \) to find \( E_{\text{final}} = K_f + U_f \).
04

Determine the Reduction in Mechanical Energy

Subtract the final mechanical energy from the initial: \[ \Delta E_{\text{mec}} = E_{\text{final}} - E_{\text{initial}} \] Since we're looking for the reduction due to air drag, we want the positive value of this difference: \[ \Delta E_{\text{drag}} = |\Delta E_{\text{mec}}| \] This gives the reduction in energy attributed to air drag.

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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 a form of energy that an object possesses due to its motion. It is defined as \( K = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
  • The faster an object moves, the greater its kinetic energy.
  • Kinetic energy depends on both mass and speed, meaning heavier and faster objects have more kinetic energy.

In the context of the Frisbee, when it鈥檚 thrown with an initial speed of 12 m/s, it's endowed with a certain amount of kinetic energy. As the Frisbee rises to a height of 2.1 m, its speed decreases to 10.5 m/s, causing a decrease in kinetic energy. This loss of kinetic energy is partly transferred to potential energy due to its gain in height and partly lost to air drag, which is the main focus of this exercise.
Potential Energy
Potential energy is the energy held by an object because of its position relative to other objects. For our purposes here, we're looking at gravitational potential energy, which can be described with the equation \( U = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity \( (9.8 \, \mathrm{m/s}^2) \), and \( h \) is the height.
  • Potential energy increases as an object is lifted higher off the ground.
  • This form of energy is "stored" energy, which can be converted back to kinetic energy.

In our Frisbee example, the initial potential energy is calculated at a height of 1.1 m. As the Frisbee ascends to a height of 2.1 m, its potential energy increases. This increase in height contributes to why some kinetic energy is converted to potential energy during the upward motion.
Air Drag
Air drag, also known as air resistance, is a force that opposes the motion of an object through the air. It can significantly affect the movement and energy of objects like a flying Frisbee.
  • Air drag acts in the opposite direction to the motion, slowing down moving objects.
  • The effect of air drag is more pronounced at higher speeds and affects objects with larger surface areas.

For our Frisbee scenario, air drag reduces the mechanical energy of the Frisbee-Earth system as it travels upward. While part of the kinetic energy is converted to potential energy, the portion of energy lost is attributed to the work done by air drag. Calculating the reduction in energy (\( \Delta E_{\text{drag}} \)) gives a measure of this energy loss, highlighting how non-conservative forces like air drag impact real-world motion.

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Most popular questions from this chapter

107 The only force acting on a particle is conservative force \(\vec{F}\). If the particle is at point \(A\), the potential energy of the system associated with \(\vec{F}\) and the particle is \(40 \mathrm{~J}\). If the particle moves from point \(A\) to point \(B\), the work done on the particle by \(\vec{F}\) is \(+25 \mathrm{~J}\). What is the potential energy of the system with the particle at \(B\) ?

A spring with a spring constant of \(3200 \mathrm{~N} / \mathrm{m}\) is initially stretched until the elastic potential energy of the spring is \(1.44 \mathrm{~J} .\) \((U=0\) for the relaxed spring.) What is \(\Delta U\) if the initial stretch is changed to (a) a stretch of \(2.0 \mathrm{~cm},(\mathrm{~b})\) a compression of \(2.0 \mathrm{~cm}\), and (c) a compression of \(4.0 \mathrm{~cm}\) ?

A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile-Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

At a certain factorv, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. 8-64). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400\). After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

The magnitude of the gravitational force between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$ F(x)=G \frac{m_{1} m_{2}}{x^{2}} $$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the corresponding potential energy function \(U(x)\) ? Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\) and that \(x\) is positive. (b) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=x_{1}+d ?\)
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