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A \(50.0-\mathrm{kg}\) projectile is fired at an angle of \(30.0^{\circ}\) above the horizontal with an initial speed of \(1.20 \times 10^{2} \mathrm{~m} / \mathrm{s}\) from the top of a cliff \(142 \mathrm{~m}\) above level ground, where the ground is taken to be \(y=0\). (a) What is the initial total mechanical energy of the projectile? (b) Suppose the projectile is traveling \(85.0 \mathrm{~m} / \mathrm{s}\) at its maximum height of \(y=427 \mathrm{~m}\). How much work has been done on the projectile by air friction? (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

Short Answer

Expert verified
First, calculate the initial total mechanical energy of the projectile as the sum of its kinetic and potential energy. Then, using the work-energy theorem, you can find out the work done by air friction by finding the difference in mechanical energy at the maximum height and at the initial height. Finally, considering the work done by air friction while going down, one can calculate the final speed of the body just before hitting the ground by equating the final mechanical energy (which is equal to the kinetic energy at ground level) to the kinetic energy expression and solving for velocity.

Step by step solution

01

Calculating initial total mechanical energy

The initial total mechanical energy of the projectile is the sum of its kinetic and potential energy. The initial kinetic energy (KE) can be calculated as: \( KE = 0.5 * mass * velocity^2 \). Here, the mass is 50.0 kg and velocity is \(1.20 x 10^{2} m/s \). The initial potential energy (PE) can be calculated as \( PE = mass * g * height \). The height here from the cliff top to ground is 142m and g is the acceleration due to gravity, which is approximately 9.81 \( m/s^2 \).
02

Calculating the work done by air friction

As per the work-energy theorem, the work done on the system is equal to its change in total mechanical energy. We are given that at the maximum height (427 m), the speed of the projectile reduces to 85.0 m/s. So, we'll first calculate the total mechanical energy at this point. The kinetic and potential energy can respectively be calculated as discussed in step 1 using these new values. The work done by air friction is then the difference between this total mechanical energy and the initial total mechanical energy.
03

Determining the speed of the projectile before hitting the ground

We are given that air friction does one-and-half times as much work on the projectile when it's going down as it did when it was going up. So, we multiply the work done by air friction in step 2 by 2.5 (1 for going up and 1.5 for going down). We then subtract this total work done by air friction from the initial total mechanical energy to get the final mechanical energy just before hitting the ground. To get the speed of the projectile, we set this equal to the kinetic energy (since the potential energy at ground level is zero) and solve for velocity.

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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 a central concept in physics that helps us understand how energy moves through systems like projectiles. In the context of projectile motion, mechanical energy is the sum of two types of energy: kinetic and potential.
In our example, the initial mechanical energy of the projectile is calculated as the sum of its initial kinetic energy and potential energy.
To find the total mechanical energy, use the formula:
  • Kinetic energy (KE): \( KE = 0.5 \times \text{mass} \times \text{velocity}^2 \)
  • Potential energy (PE): \( PE = \text{mass} \times g \times \text{height} \)
Where:
  • \( g = 9.81 \, \text{m/s}^2 \), the acceleration due to gravity.
  • For the cliff problem, height is 142 m.
By calculating these energies, you can find the total mechanical energy available to the projectile before other forces, like air friction, come into play.
Kinetic Energy
Kinetic energy represents the energy a projectile possesses due to its motion.
When a projectile is fired, its initial speed contributes to its kinetic energy and is given by the formula:
  • \( KE = 0.5 \times \text{mass} \times \text{velocity}^2 \)
For our example, the mass is 50.0 kg, and the initial velocity is 120 m/s. Kinetic energy is a vital component because it conveys how much energy due to motion is available at any point in the projectile’s path. Note how this energy changes when the velocity of the projectile changes, such as when it reaches its maximum height. Kinetic energy decreases as the projectile slows down due to air friction and gravitational pull, emphasizing that motion significantly affects energy content.
Potential Energy
Potential energy is the energy stored in a system due to its position in a gravitational field.
For our projectile, the potential energy is initially due to its height above the ground:
\( PE = \text{mass} \times g \times \text{height} \)
In our problem, the projectile is launched from a cliff 142 meters above the ground. This height gives the projectile potential energy even when it is stationary. As the projectile rises, potential energy increases, reaching its maximum when the projectile is at its highest point. Nevertheless, potential energy plays a crucial role since as it decreases while the projectile is descending, it transforms into kinetic energy. This transformation emphasizes the interplay between kinetic and potential energy in maintaining total mechanical energy.
Work-Energy Theorem
The work-energy theorem is a fundamental principle connecting the work done on an object to its energy changes. It states that the work done on an object is equal to its change in kinetic energy.
In terms of projectile motion, this theorem is crucial for determining how external forces, like air friction, influence the projectile’s total energy.
The formula used is:
  • \( \text{Work} = \Delta \, \text{Kinetic Energy} \)
In our scenario, calculating the work done by air friction involves determining the difference between the total mechanical energy initially and when the projectile is at its highest point. The works calculated help in understanding how much the air resistance offsets the projectile’s trajectory and energy.
Air Friction
Air friction, or air resistance, is the force acting opposite to the relative motion of an object moving with respect to a surrounding fluid, which, in this case, is air.
It affects the projectile’s velocity and trajectory, causing a loss in mechanical energy. In our exercise, air friction does work on the projectile, decreasing its velocity. Initially, it reduces the velocity from 120 m/s to 85 m/s at its maximum altitude. Hence, the work done by air friction can be calculated as the difference in total mechanical energy at these points.
Furthermore, it’s noteworthy that as the projectile descends, air friction does even more work, reducing its final speed before impact, illustrating how air friction must be considered in real-world physics to ensure precise calculations.

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Most popular questions from this chapter

A ski jumper starts from rest \(50.0 \mathrm{~m}\) above the ground on a frictionless track and flies off the track at an angle of \(45.0^{\circ}\) above the horizontal and at a height of \(10.0 \mathrm{~m}\) above the level ground. Neglect air resistance. (a) What is her speed when she leaves the track? (b) What is the maximum altitude she attains after leaving the track? (c) Where does she land relative to the end of the track?

A roller-coaster car of mass \(1.50 \times 10^{3} \mathrm{~kg}\) is initially at the top of a rise at point (A). It then moves \(35.0 \mathrm{~m}\) at an angle of \(50.0^{\circ}\) below the horizontal to a lower point (B). (a) Find both the potential energy of the system when the car is at points (A) and (B) and the change in potential energy as the car moves from point (A) to point (B), assuming \(y=0\) at point. (B). (b) Repeat part (a), this time choosing \(y=0\) at point (C), which is another \(15.0 \mathrm{~m}\) down the same slope from point (B).

M A 2 100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls \(5.00 \mathrm{~m}\) before coming into contact with the top of the beam, and it drives the beam \(12.0 \mathrm{~cm}\) farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

A skier starts from rest at the top of a hill that is inclined \(10.5^{\circ}\) with respect to the horizontal. The hillside is \(200 \mathrm{~m}\) long, and the coefficient of friction between snow and skis is \(0.0750\). At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

In 1990 Walter Arfeuille of Belgium lifted a \(281.5-\mathrm{kg}\) object through a distance of \(17.1 \mathrm{~cm}\) using only his teeth. (a) How much work did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?

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