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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

Identify Initial and Final Velocities

The initial speed of the skier is given as \( v_i = 24 \mathrm{~m/s} \) at an angle of \( \theta = 25^\circ \) above the horizontal. The final speed of the skier when landing is \( v_f = 22 \mathrm{~m/s} \).
02

Calculate Initial and Final Kinetic Energy

The initial kinetic energy \( KE_i \) is given by \( \frac{1}{2} m v_i^2 \). So, \( KE_i = \frac{1}{2} \times 60 \mathrm{~kg} \times (24 \mathrm{~m/s})^2 = 17280 \mathrm{~J} \). The final kinetic energy \( KE_f \) is \( \frac{1}{2} m v_f^2 = \frac{1}{2} \times 60 \mathrm{~kg} \times (22 \mathrm{~m/s})^2 = 14520 \mathrm{~J} \).
03

Calculate Change in Potential Energy

The change in potential energy due to vertical movement is \( \Delta PE = mgh \), where \( h = 14 \mathrm{~m} \) and \( g = 9.8 \mathrm{~m/s^2} \). So, \( \Delta PE = 60 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} \times 14 \mathrm{~m} = 8232 \mathrm{~J} \).
04

Calculate Initial Energy and Final Energy

The initial mechanical energy \( E_i \) is the sum of initial kinetic energy and initial potential energy (assuming potential energy at the base is zero). \( E_i = KE_i + PE_i = 17280 \mathrm{~J} \) (since the height is considered from the jump point, initial PE is essentially zero here).The final mechanical energy \( E_f \) becomes the sum of \( KE_f \) and \(-8232 \mathrm{~J}\) for the potential energy change. Thus, \( E_f = 14520 \mathrm{~J} - 8232 \mathrm{~J} = 6288 \mathrm{~J} \).
05

Determine the Reduction in Mechanical Energy

The reduction in mechanical energy due to air drag is the difference \( \Delta E = E_i - E_f = 17280 \mathrm{~J} - 6288 \mathrm{~J} = 10992 \mathrm{~J} \).

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Key Concepts

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

Kinetic Energy
When discussing the motion of the skier, kinetic energy plays a pivotal role. Kinetic energy (\( KE \)) is the energy that an object possesses due to its motion. It is determined by the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the skier, and \( v \) is the velocity at which the skier is moving.
In our exercise, the skier starts with an initial speed of 24 m/s, leading to an initial kinetic energy calculation as follows:
  • \( KE_i = \frac{1}{2} \times 60 \mathrm{~kg} \times (24 \mathrm{~m/s})^2 = 17280 \mathrm{~J} \)
Upon landing, the skier's speed reduces to 22 m/s, changing the kinetic energy to:
  • \( KE_f = \frac{1}{2} \times 60 \mathrm{~kg} \times (22 \mathrm{~m/s})^2 = 14520 \mathrm{~J} \)
This reduction in kinetic energy highlights how velocity plays a significant role in determining the energy of moving objects like our skier.
Potential Energy
Potential energy (\( PE \)) refers to the energy held by an object due to its position relative to other objects. In the context of our skier, it is primarily gravitational potential energy. The formula to determine potential energy is:
  • \( \Delta PE = mgh \)
where \( m \) is the mass of the skier, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height difference the skier experiences.
For the skier, the change in height upon landing is 14 m below the end of the ramp, which results in a potential energy change calculated as:
  • \( \Delta PE = 60 \mathrm{~kg} \times 9.8 \mathrm{~m/s}^2 \times 14 \mathrm{~m} = 8232 \mathrm{~J} \)
In this scenario, the decrease in potential energy indicates the work done by gravity as the skier descends.
Energy Loss due to Air Drag
Air drag, or air resistance, acts against the motion of the skier, causing energy loss. This effect is part of the reason why the skier's velocity decreases as they descend, alongside gravitational forces. By calculating the total mechanical energy change, we can quantify the energy lost owing to air drag.
The initial total mechanical energy (\( E_i \)) is equivalent to the sum of the initial kinetic energy and initial potential energy. In this example:
  • \( E_i = KE_i + PE_i = 17280 \mathrm{~J} \)
The final mechanical energy (\( E_f \)) is the combination of the final kinetic energy and the negative potential energy change.
  • \( E_f = 14520 \mathrm{~J} - 8232 \mathrm{~J} = 6288 \mathrm{~J} \)
Therefore, the energy reduced due to air drag can be calculated by:
  • \( \Delta E = E_i - E_f = 17280 \mathrm{~J} - 6288 \mathrm{~J} = 10992 \mathrm{~J} \)
This reduction highlights how air resistance plays a crucial role in energy dynamics by influencing the skier's final speed and overall energy during the jump.

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Most popular questions from this chapter

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N}\), the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\) with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\).

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. 8-35) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

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?

A chain is held on a frictionless table with onefourth of its length hanging over the edge. If the chain has length \(L=28 \mathrm{~cm}\) and mass \(m=0.012 \mathrm{~kg}\), how much work is required to pull the hanging part back onto the table?

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