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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 \(27 \mathrm{~m} / \mathrm{s}\) 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 15,582 J due to air drag.

Step by step solution

01

Calculate Initial Kinetic Energy

The initial kinetic energy \( KE_i \) can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the skier and \( v \) is the velocity. The initial velocity \( v_i \) is given as 27 m/s. Substitute the values to find \( KE_i \): \( KE_i = \frac{1}{2} \times 60 \times 27^2 \). This simplifies to \( KE_i = 21,870 \mathrm{~J} \).
02

Calculate Initial Potential Energy

The initial potential energy \( PE_i \) is zero because the ski-jump is the reference point (\( h = 0 \)). Thus, \( PE_i = 0 \).
03

Calculate Final Kinetic Energy

The final kinetic energy \( KE_f \) is calculated with the final speed of 22 m/s: \( KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 60 \times 22^2 \). Simplifying gives \( KE_f = 14,520 \mathrm{~J} \).
04

Calculate Final Potential Energy

The final potential energy \( PE_f \) is given by \( mgh \), where \( g = 9.8 \mathrm{~m/s^2} \) and \( h = -14 \mathrm{~m} \) (since he descends by 14 m). Thus, \( PE_f = 60 \times 9.8 \times -14 = -8,232 \mathrm{~J} \).
05

Determine Initial Mechanical Energy

The initial mechanical energy \( E_i \) is the sum of \( KE_i \) and \( PE_i \): \( E_i = 21,870 + 0 = 21,870 \mathrm{~J} \).
06

Determine Final Mechanical Energy

The final mechanical energy \( E_f \) is the sum of \( KE_f \) and \( PE_f \): \( E_f = 14,520 - 8,232 = 6,288 \mathrm{~J} \).
07

Calculate the Reduction in Mechanical Energy Due to Air Drag

The reduction in mechanical energy due to air drag is the difference between initial and final mechanical energy: \( \Delta E = E_i - E_f = 21,870 - 6,288 = 15,582 \mathrm{~J} \). Thus, the mechanical energy is reduced by 15,582 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, plays a crucial role in the motion of objects through the air. It acts as a force that opposes the movement of an object. In the case of our skier, air drag reduces their speed as they descend, impacting their overall energy. This resistance arises due to friction between the skier and the air molecules.
  • Air drag is proportional to the velocity of the object squared, meaning faster objects experience greater drag.
  • It's dependent on the surface area and shape of the object; smoother and smaller areas tend to face less resistance.
  • This force is vital in real-world scenarios like skiing, where it significantly affects an athlete's performance and mechanical energy.
Understanding air drag allows us to explain why the skier's final speed was lower than the initial speed despite descending the ramp.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. For our skier, both their initial and final kinetic energies are crucial in understanding the total mechanical energy of the system. Kinetic energy is given by the formula:\[KE = \frac{1}{2}mv^2\]where \( m \) is the mass of the object and \( v \) is its velocity.
  • At the beginning of the jump, the skier has a high kinetic energy due to their high speed.
  • Even though velocities were calculated as 27 m/s and 22 m/s, the significant difference causes a decrease in kinetic energy.
  • This change provides insight into the energy lost due to factors like air drag.
Observing how kinetic energy changes helps us understand the effects of forces on a moving object.
Potential Energy
Potential energy is the stored energy of an object due to its position or height. In the ski-jump problem, it gives insights into energy changes as the skier descends. Potential energy is calculated using the formula:\[PE = mgh\]where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is height.
  • Initially, potential energy was set to zero, as the skier was at the reference height of the ramp.
  • As the skier descends 14 meters, their potential energy becomes negative, indicating a decrease due to a lower position.
  • This decrease reflects the loss of stored energy as kinetic energy transforms during the movement.
By observing these potential energy changes, one can understand how height variations influence energy storage.
Energy Conservation
The principle of energy conservation states that the total mechanical energy in an isolated system remains constant in the absence of external forces like friction or air drag. However, in real-world applications like skiing, external forces cause energy transformations and reductions.
  • Initially, the skier had a certain amount of mechanical energy, a sum of kinetic and potential energies.
  • As they proceeded, the mechanical energy transformed, with some lost to air drag and friction.
  • The significant reduction in the skier's total mechanical energy by 15,582 J shows the impact of non-conservative forces.
This principle helps explain the differences between initial and final mechanical energy, emphasizing the real-world effects of such forces.
Physics Problem Solving
Approaching physics problems systematically is vital for understanding and solving complex scenarios. It involves breaking the problem into smaller steps and analyzing each component separately.
  • Start by identifying known quantities and required outcomes, such as masses, velocities, and heights.
  • Calculate individual energies, both kinetic and potential, at important positions in the problem.
  • Sum these values for total mechanical energy and identify where changes occur, particularly due to external forces like air drag.
By following a structured problem-solving approach, like the one used in the ski-jump problem, students can gain insights into physical concepts and accurately determine energy variations.

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Most popular questions from this chapter

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(150 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.36 ?\)

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(11.5 \mathrm{~m}\) high. The snowball's initial velocity is \(16.0 \mathrm{~m} / \mathrm{s}\), directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

A loaded truck of mass \(3000 \mathrm{~kg}\) moves on a level road at a constant speed of \(6.000 \mathrm{~m} / \mathrm{s}\). The frictional force on the truck from the road is \(1000 \mathrm{~N}\). Assume that air drag is negligible. (a) How much work is done by the truck engine in \(10.00 \mathrm{~min}\) ? (b) After \(10.00 \mathrm{~min}\), the truck enters a hilly region whose inclination is \(30^{\circ}\) and continues to move with the same speed for another \(10.00 \mathrm{~min}\). What is the total work done by the engine during that period against the gravitational force and the frictional force? (c) What is the total work done by the engine in the full \(20 \mathrm{~min}\) ?

Tarzan, who weighs \(688 \mathrm{~N}\), swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long (Fig. 8-28). From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\) (a) Does the vine break? (b) If it does not, what is the greatest force on it during the swing? If it does, at what angle with the vertical does it break?
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