/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A \(60 \mathrm{~kg}\) skier leav... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(24 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of \(22 \mathrm{~m} / \mathrm{s},\) landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

Short Answer

Expert verified
The mechanical energy is reduced by 10992 J due to air drag.

Step by step solution

01

Set up the Problem

We need to determine the change in mechanical energy of the skier-Earth system, with mechanical energy consisting of potential and kinetic energy. The reduction is due to air drag. We will calculate the mechanical energy at the launch and landing points, then find the difference.
02

Calculate Initial Mechanical Energy

The initial mechanical energy consists of kinetic energy (KE) and gravitational potential energy (PE). The initial kinetic energy is calculated as \( KE_i = \frac{1}{2}mv_i^2 \), where \( m = 60 \, \text{kg} \) and \( v_i = 24 \, \text{m/s} \). Initial potential energy is \( PE_i = mgh_i \), where the vertical height \( h_i \) is zero, since it's the reference point: \( PE_i = 0 \).
03

Compute Initial Kinetic Energy

Compute the initial kinetic energy:\[KE_i = \frac{1}{2} \times 60 \, \text{kg} \times (24 \, \text{m/s})^2 = 17280 \, \text{J}.\]
04

Calculate Final Mechanical Energy

The final mechanical energy also has components of kinetic and potential energy. The final kinetic energy is \( KE_f = \frac{1}{2}mv_f^2 \) with \( v_f = 22 \, \text{m/s} \). The final potential energy \( PE_f = mgh_f \), with \( h_f = -14 \, \text{m} \).
05

Compute Final Kinetic Energy

Calculate the final kinetic energy:\[KE_f = \frac{1}{2} \times 60 \, \text{kg} \times (22 \, \text{m/s})^2 = 14520 \, \text{J}.\]
06

Compute Final Potential Energy

Calculate the final potential energy:\[PE_f = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times (-14 \, \text{m}) = -8232 \, \text{J}.\]
07

Find Total Initial Mechanical Energy

The total initial mechanical energy is the sum of initial kinetic and potential energies: \[E_{i} = KE_i + PE_i = 17280 \, \text{J} + 0 = 17280 \, \text{J}.\]
08

Find Total Final Mechanical Energy

The total final mechanical energy is the sum of final kinetic and potential energies: \[E_{f} = KE_f + PE_f = 14520 \, \text{J} + (-8232 \, \text{J}) = 6288 \, \text{J}.\]
09

Calculate the Reduction in Mechanical Energy

The reduction in mechanical energy is the difference between the initial and final total mechanical energies:\[\Delta E = E_i - E_f = 17280 \, \text{J} - 6288 \, \text{J} = 10992 \, \text{J}.\]
10

Conclusion

Due to air drag, the mechanical energy of the skier-Earth system is reduced by \( 10992 \, \text{J} \).

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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 an object has due to its motion. It depends on two main factors: the mass of the object and its velocity. For any moving object, you can calculate the kinetic energy using the formula:
  • \( KE = \frac{1}{2}mv^2 \)
In this formula, \( m \) stands for mass, measured in kilograms, and \( v \) stands for velocity, measured in meters per second. The kinetic energy is measured in joules (J).
In our exercise, at the start, the skier has a high velocity of 24 m/s, giving them a significant amount of kinetic energy. Before launching off the ramp, their kinetic energy was calculated to be 17280 J. Even though friction and other forces such as air drag can reduce speed, it is the starting point of the skier's mechanical journey.
Potential Energy
Potential energy refers to the stored energy associated with the position of an object. The most common type, gravitational potential energy, occurs due to an object's height above a reference point. This can be calculated using the formula:
  • \( PE = mgh \)
Where \( m \) represents mass, \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and \( h \) represents height above the reference point.
In the skier's scenario, initially launching from the ramp, the potential energy is zero as it's the reference height. However, when the skier lands 14 meters below the initial jump, it creates negative potential energy, meaning they have descended from their original position. This is calculated as \(-8232 \, J\) when they hit the ground.
Air Drag
Air drag, also known as air resistance, is the force that opposes an object's motion through the air. It's a type of friction that objects encounter when moving. Air drag can considerably reduce an object's velocity, impacting the mechanical energy of a moving system.
As the skier moved through the air, air drag decreased their speed from 24 m/s at launch to 22 m/s upon landing. This reduction in speed is a clear indication of energy loss due to the air drag effect. Air drag does work against the skier’s motion, turning some kinetic energy into thermal energy, thus not available for the skier’s mechanical movement.
Energy Conservation
The principle of energy conservation states that energy in a closed system remains constant. However, in reality, external forces such as air drag can dispense energy out of the system, converting mechanical energy into other forms such as heat.

In the scenario of our skier, although initial mechanical energy (potential plus kinetic) was tallied at launch, air drag changed the energy landscape. Initially starting with 17280 J of energy and landing with only 6288 J, energy was lost.
  • The loss, calculated as 10992 J, highlights the effect of external forces working against the skier.
  • This energy reduction demonstrates how systems in practice deviate slightly from ideal conservation principles, primarily due to real-world forces such as air drag.
Understanding these principles helps us realize that while energy cannot be created or destroyed, it can certainly change form within different contexts.

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Most popular questions from this chapter

Each second, \(1200 \mathrm{~m}^{3}\) of water passes over a waterfall \(100 \mathrm{~m}\) high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

Approximately \(5.5 \times 10^{6} \mathrm{~kg}\) of water falls \(50 \mathrm{~m}\) over Niagara Falls each second. (a) What is the decrease in the gravitational potential energy of the water-Earth system each second? (b) If all this energy could be converted to electrical energy (it cannot be), at what rate would electrical energy be supplied? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg} .\) ) (c) If the electrical energy were sold at 1 cent \(/ \mathrm{kW} \cdot \mathrm{h},\) what would be the yearly income?

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=-4 x e^{-x / 4} \mathrm{~J} $$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m},\) and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0 ?\)

An automobile with passengers has weight \(16400 \mathrm{~N}\) and is moving at \(113 \mathrm{~km} / \mathrm{h}\) when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of \(8230 \mathrm{~N}\). Find the stopping distance.

A \(20 \mathrm{~kg}\) object is acted on by a conservative force given by \(F=-3.0 x-5.0 x^{2},\) with \(F\) in newtons and \(x\) in meters. Take the potential energy associated with the force to be zero when the object is at \(x=0 .\) (a) What is the potential energy of the system associated with the force when the object is at \(x=2.0 \mathrm{~m} ?\) (b) If the object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis when it is at \(x=5.0 \mathrm{~m},\) what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be \(-8.0 \mathrm{~J}\) when the object is at \(x=0 ?\)

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