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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 air drag causes a change of -0.076 J in the mechanical energy.

Step by step solution

01

Calculate Initial Mechanical Energy

First, we need to calculate the initial mechanical energy of the system. The initial mechanical energy consists of the initial kinetic energy and the initial potential energy. The initial kinetic energy is given by: \[ K_i = \frac{1}{2} m v^2 \] where \( m = 0.50 \text{ kg} \) and \( v = 4.00 \text{ m/s} \). Plugging in the values, \[ K_i = \frac{1}{2} \times 0.50 \times (4.00)^2 = 4.00 \text{ J} \]The initial potential energy \( U_i \) at the surface (height = 0) is 0, so the total initial mechanical energy \( E_i = K_i + U_i = 4.00 \text{ J} \).
02

Calculate Final Mechanical Energy at Maximum Height

Next, calculate the mechanical energy at the maximum height. At the maximum height, the speed is 0, so the kinetic energy is 0. The potential energy at maximum height is given by: \[ U_f = mgh \] where \( h = 0.80 \text{ m} \) and \( g = 9.81 \text{ m/s}^2 \). Plugging in the values, \[ U_f = 0.50 \times 9.81 \times 0.80 = 3.924 \text{ J} \]Thus, the final mechanical energy \( E_f = U_f = 3.924 \text{ J} \).
03

Calculate the Change in Mechanical Energy Due to Air Drag

Finally, find the change in mechanical energy due to air drag, which is the difference between the initial and final mechanical energies: \[ \Delta E = E_f - E_i \] Substituting the known values, \[ \Delta E = 3.924 \text{ J} - 4.00 \text{ J} = -0.076 \text{ J} \]The negative sign indicates a loss in mechanical energy due 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 refers to the energy that an object possesses due to its motion. When a banana is thrown upwards, it initially moves with a certain speed, giving it kinetic energy. This energy can be calculated using the formula:
  • \( K = \frac{1}{2} m v^2 \)
where \(m\) is the mass of the object and \(v\) is its velocity.
In our exercise, the banana has a mass of 0.50 kg and is thrown with a speed of 4.00 m/s. Plugging these values into the formula, we find the initial kinetic energy to be 4.00 J (joules).
Understanding kinetic energy is crucial because it allows us to predict how the object's energy will change as it moves, especially when other forces, like air drag, come into play.
Potential Energy
Potential energy is the energy stored in an object due to its position in a force field, typically gravitational. When the banana is thrown upwards, it climbs against gravity, gaining potential energy. This energy can be calculated with:
  • \( U = mgh \)
where \(m\) is mass, \(g\) is the acceleration due to gravity (9.81 m/s\(^2\)), and \(h\) is the height.
In this scenario, the banana reaches a maximum height of 0.80 m. At this peak, its velocity is zero, so all its initial kinetic energy has been converted to potential energy. The calculation reveals a potential energy of 3.924 J.
This concept helps us understand energy transformations that occur as an object moves through different heights.
Air Drag
Air drag, also known as air resistance, is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. For the case of the banana, air drag causes a loss of mechanical energy as it ascends.
  • Mechanical energy is the sum of kinetic and potential energy.
  • In an ideal scenario without air resistance, mechanical energy remains constant.
However, in our exercise, the initial mechanical energy is 4.00 J and the energy at the height of 0.80 m is 3.924 J.
The difference, -0.076 J, signifies the energy lost due to air drag. This loss is essential to consider, as it impacts how high and fast objects can move through the air.
Understanding air drag is vital when studying real-life scenarios, enabling us to make accurate predictions about the motion and energy changes in objects.

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Most popular questions from this chapter

A \(0.42 \mathrm{~kg}\) shuffleboard disk is initially at rest when a player uses a cue to increase its speed to \(4.2 \mathrm{~m} / \mathrm{s}\) at constant acceleration. The acceleration takes place over a \(2.0 \mathrm{~m}\) distance, at the end of which the cue loses contact with the disk. Then the disk slides an additional \(12 \mathrm{~m}\) before stopping. Assume that the shuffleboard court is level and that the force of friction on the disk is constant. What is the increase in the thermal energy of the disk-court system (a) for that additional \(12 \mathrm{~m}\) and \((\mathrm{b})\) for the entire \(14 \mathrm{~m}\) distance? (c) How much work is done on the disk by the cue?

A rope is used to pull a \(3.57 \mathrm{~kg}\) block at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor. The force on the block from the rope is \(7.68 \mathrm{~N}\) and directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

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?

Shows a pendulum of length \(L=1.25 \mathrm{~m}\). Its bob (which effectively has all the mass) has speed \(v_{0}\) when the cord makes an angle \(\theta_{0}=40.0^{\circ}\) with the vertical. (a) What is the speed of the bob when it is in its lowest position if \(v_{0}=8.00 \mathrm{~m} / \mathrm{s} ?\) What is the least value that \(v_{0}\) can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if \(\theta_{0}\) is increased by a few degrees?

The string is \(L=120 \mathrm{~cm}\) long, has a ball attached to one end, and is fixed at its other end. A fixed peg is at point \(P\). Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distance \(d \mathrm{ex}\) ceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)
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