/*! 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] | 91影视

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Chapter 8: Problem 50

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

Understand the Problem

We need to calculate the change in mechanical energy due to air drag as the skier jumps and lands. The mechanical energy consists of both kinetic and potential energy.
02

Calculate Initial Mechanical Energy

First, calculate the initial potential energy (PE_initial) and kinetic energy (KE_initial) at the top of the jump. For potential energy: \[ PE_{initial} = mgh = 0 \] since the height at the starting point is zero.For kinetic energy: \[ KE_{initial} = \frac{1}{2}mv^2 = \frac{1}{2} imes 60 imes 24^2 = 17280 \, \text{J} \] These give us the initial mechanical energy: \[ ME_{initial} = PE_{initial} + KE_{initial} = 0 + 17280 = 17280 \, \text{J} \]
03

Calculate Final Mechanical Energy

Next, calculate the final potential energy (PE_final) and kinetic energy (KE_final) at the landing point. For potential energy: \[ PE_{final} = mgh = 60 \times 9.8 \times (-14) = -8232 \, \text{J} \]For kinetic energy:\[ KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2} \times 60 \times 22^2 = 14520 \, \text{J} \] Then, calculate the final mechanical energy:\[ ME_{final} = PE_{final} + KE_{final} = -8232 + 14520 = 6288 \, \text{J} \]
04

Calculate Change in Mechanical Energy

Finally, find the reduction in mechanical energy due to air drag by subtracting the final mechanical energy from the initial mechanical energy: \[ \Delta ME = ME_{initial} - ME_{final} = 17280 - 6288 = 10992 \, \text{J} \]
05

Conclusion

The mechanical energy of the skier-Earth system is reduced by 10992 J due to air drag.

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Key Concepts

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

Air Drag
Air drag, also known as air resistance, is a force that opposes the motion of an object through the air. It occurs due to collisions between the object's surface and air molecules. When a skier moves through the air, this force causes a decrease in speed over time.
Air drag depends on several factors, such as:
  • The speed of the skier鈥攈igher speeds generally increase the drag force.
  • The shape and surface area of the skier鈥攖he more streamlined and smaller reduces drag.
  • The density of the air, which can vary with altitude and temperature.
Air drag is not negligible in scenarios like skiing, especially during jumps where high velocities are involved. It causes a reduction in the total mechanical energy of the skier-Earth system, as evidenced by the skier's lower speed upon landing compared to launching.
Potential Energy
Potential energy is the energy stored within an object due to its position relative to other objects. For a skier, potential energy changes based on their height above the start point.
In our scenario, potential energy is calculated using:
  • The mass of the skier, which is a constant at 60 kg.
  • The gravitational force, approximately 9.8 m/s虏 on Earth.
  • The height difference, which is 14 m below the initial point in this exercise.
The potential energy decrease of the skier at landing鈥攃alculated as a negative value鈥攔eflects how much energy is no longer stored as height above the ground, highlighting significant energy conversion in the system through the jump.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a crucial aspect of mechanical energy, involving factors like mass and speed.
From the problem's perspective, we calculate kinetic energy using:
  • The skier's mass, remaining constant at 60 kg.
  • The speed of the skier, initially 24 m/s and reducing to 22 m/s when landing.
The formula is: \[ KE = \frac{1}{2} m v^2 \]The skier's initial kinetic energy was higher due to greater speed, reducing as the skier lands. This reduction accounts for a portion of the overall energy lost during the jump, with air drag contributing to the decreasing speed.
Energy Conservation
Energy conservation is a principle stating that the total energy in an isolated system remains constant鈥攊t can neither be created nor destroyed, only transformed.
In the skier's case, the mechanical energy consisted of both kinetic and potential energy. Ideally, the sum of these energies would remain the same. However, due to air drag, some energy converted into non-mechanical forms like heat and sound.
The striking fact from the exercise is that the mechanical energy of the skier-Earth system reduces by 10992 J due to energy conversion. Understanding energy conservation allows us to recognize how forces like air drag can impact a system鈥檚 energy balance. Despite all transformations, the total energy accounting for all forms remains unchanged; it underscores the importance of identifying and understanding energy loss mechanisms in practical scenarios.

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Most popular questions from this chapter

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.

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?(\mathrm{c})\) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m} .\) The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0 \mathrm{~cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

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?

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?
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