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Chapter 8: Problem 97

A \(0.50 \mathrm{~kg}\) banana is thrown directly upward with an initial speed of \(4.00 \mathrm{~m} / \mathrm{s}\) and reaches a maximum height of \(0.80 \mathrm{~m}\). What change does air drag cause in the mechanical energy of the banana-Earth system during the ascent?

Short Answer

Expert verified
The change in mechanical energy due to air drag is -0.076 J.

Step by step solution

01

Calculate Initial Mechanical Energy

First, we calculate the initial mechanical energy of the banana when it is thrown upward. It consists of the kinetic and gravitational potential energy at the initial point. The kinetic energy (KE) is given by \( KE_i = \frac{1}{2}mv^2 \), where \( m = 0.50 \, \text{kg} \) and \( v = 4.00 \, \text{m/s} \). So, \( KE_i = \frac{1}{2} \times 0.50 \times (4.00)^2 = 4.00 \, \text{J} \). The initial potential energy (PE) is zero because we consider the starting point as the reference level (height = 0). Therefore, the initial mechanical energy is \( 4.00 \, \text{J} \).
02

Calculate Final Mechanical Energy

Next, we calculate the mechanical energy at the maximum height. At this point, the kinetic energy is zero because the banana's speed is zero. The potential energy (PE) is given by \( PE_f = mgh \), where \( h = 0.80 \, \text{m} \) and \( g = 9.81 \, \text{m/s}^2 \). So, \( PE_f = 0.50 \times 9.81 \times 0.80 = 3.924 \, \text{J} \). Thus, the final mechanical energy is \( 3.924 \, \text{J} \).
03

Determine Change in Mechanical Energy

The change in mechanical energy due to air drag can be determined by subtracting the final mechanical energy from the initial mechanical energy. So, the change is \( \Delta E = 4.00 - 3.924 = 0.076 \, \text{J} \). This value represents the mechanical energy lost to air drag during the ascent.

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Key Concepts

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

Mechanical Energy
Mechanical energy is the sum of both kinetic and potential energy in an object. It represents the total energy that an object possesses due to its motion and position in a gravitational field. In this exercise, the banana is launched upwards and encounters the forces of nature (gravity and air drag) that alter its mechanical energy.

When the banana is thrown, it starts with a certain amount of mechanical energy, expressed as the sum of its kinetic energy (from its motion) and potential energy (from its position relative to the ground). Understanding mechanical energy helps us see how different forces, like air drag, can change during an object's motion.
Kinetic Energy
Kinetic energy is the energy of motion. For the banana, its initial kinetic energy is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass and \( v \) is the velocity of the banana.

The initial kinetic energy of the banana, just after it is thrown, is determined to be \( 4.00 \text{ J} \), owing to its upward speed of \( 4.00 \text{ m/s} \). This energy represents the capability of the banana to do work against gravitational force as it moves upward. Kinetic energy is maximum when the banana is released and zero at its peak height.
Potential Energy
Potential energy is the energy held by an object due to its position relative to other objects. As the banana rises, it gains height and thus gains potential energy stored in the system.

At the top, where the banana momentarily stops, its potential energy is maximized. This is calculated with the formula:\[ PE = mgh \]where \( m \) is mass, \( g \) is the gravitational acceleration, and \( h \) is the height the banana reaches. For our banana, at a height of \( 0.80 \text{ m} \), its potential energy becomes \( 3.924 \text{ J} \). At this point, the kinetic energy is zero because the upward motion has stopped.
Air Drag
Air drag, or air resistance, is the force that opposes the motion of an object through the air. It acts to reduce the mechanical energy of the object as it moves. In the case of the banana's ascent, air drag plays a crucial role in the energy loss observed.

Although not directly calculated, the presence of air drag is evident because not all of the initial kinetic energy gets converted to potential energy. Some is lost due to the air drag overcoming the banana's upward motion. This concept helps explain why mechanical energy is lower at the peak than initially calculated.
Energy Loss
Energy loss occurs when the total mechanical energy of an object decreases, often due to external forces like air drag. In this exercise, the change in mechanical energy of the banana-Earth system is calculated to be \( 0.076 \text{ J} \).

This lost energy represents the work done by air drag on the banana. As mechanical energy dissipates, it is transformed into heat or other forms of energy, reflecting the efficiency and influences in real-world physical systems. Insight into energy loss helps understand how energy conservation can be influenced by frictional forces in everyday scenarios.

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Most popular questions from this chapter

A swimmer moves through the water at an average speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force is \(110 \mathrm{~N}\). What average power is required of the swimmer?

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\), where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\). with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and \((\mathrm{d})\) positive value of \(x\) is the potential energy equal to zero?

A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and \((\mathrm{b})\) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on (c) the mass of the ball or (d) the initial angle?

A volcanic ash flow is moving across horizontal ground when it encounters a \(10^{\circ}\) upslope. The front of the flow then travels 920 \(\mathrm{m}\) up the slope before stopping. Assume that the gases entrapped in the flow lift the flow and thus make the frictional force from the ground negligible; assume also that the mechanical energy of the front of the flow is conserved. What was the initial speed of the front of the flow?

Tarzan, who weighs \(688 \mathrm{~N}\), swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long (Fig. 8-40). From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\). (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?
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