/*! 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 22 A \(60 \mathrm{~kg}\) skier star... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
(a) Maximum height above ramp is 4.4 m. (b) Maximum height remains the same with increased weight.

Step by step solution

01

Calculate skier's velocity at the end of the ramp

Using conservation of energy, initial potential energy at height \(H\) is converted to kinetic energy at the end of the ramp. The equation is: \[ mgh = \frac{1}{2}mv^2 \]where \(m = 60\, \mathrm{kg}\), \(g = 9.8\, \mathrm{m/s^2}\), and \(h = 20\, \mathrm{m}\). Solve for \(v\):\[ v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 20} \approx 19.8\, \mathrm{m/s} \]
02

Decompose velocity into vertical and horizontal components

The skier leaves the ramp at angle \(\theta = 28^{\circ}\). Decompose the velocity \(v\) into vertical \(v_y\) and horizontal \(v_x\) components using:\[ v_y = v \sin \theta = 19.8 \sin 28^{\circ} \approx 9.3\, \mathrm{m/s} \]\[ v_x = v \cos \theta = 19.8 \cos 28^{\circ} \approx 17.5\, \mathrm{m/s} \]
03

Calculate maximum height of jump above ramp

At maximum height, the vertical velocity is zero. Using the formula for vertical motion:\[ v_{y, \text{final}}^2 = v_y^2 - 2gh \]Set \(v_{y, \text{final}} = 0\) and solve for \(h\):\[ h = \frac{v_y^2}{2g} = \frac{(9.3)^2}{2 \times 9.8} \approx 4.4\, \mathrm{m} \]
04

Understand effect of increased weight

Since energy conservation depends on height and not mass, increasing mass with a backpack does not affect the maximum height \(h\). It remains the same regardless of the skier’s weight.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy states that the total energy in a closed system remains constant. In our physics problem, a skier starts from rest and moves down a frictionless ramp. Initially, all the skier's energy is in the form of potential energy due to gravity:
  • Potential Energy = \( mgh \)
As the skier descends, this potential energy is converted into kinetic energy. Just before leaving the ramp, this energy conversion is described by the equation:
  • Kinetic Energy = \( \frac{1}{2}mv^2 \)
Since there is no friction, we can assume energy is fully conserved in this system. The relationship is expressed as:\[ mgh = \frac{1}{2}mv^2 \]Here, mass \( m \) cancels out, which simplifies our calculations. This indicates that for a frictionless surface, the skier's mass does not influence how fast they are moving just before leaving the ramp.
Projectile Motion
When the skier leaves the ramp, he becomes a projectile. Projectile motion involves analyzing the path an object takes under the influence of gravity. This motion can be split into two components:
  • Horizontal Motion
  • Vertical Motion
Each component acts independently.
For our skier:
  • The horizontal component remains constant since we ignore air resistance: \( v_x = v \cos \theta \)
  • The vertical component changes due to gravity: \( v_y = v \sin \theta \)
This dual nature of motion helps determine how far and how high the skier will travel once airborne.
Kinematics
Kinematics is the branch of physics dealing with motion without considering its causes. In the skier's jump, we calculate the maximum height attained using kinematic equations.
Specifically, when the skier reaches maximum height, the vertical velocity becomes zero. This can be calculated using the equation:
  • \( v_{y, \text{final}}^2 = v_y^2 - 2gh \)
Setting \( v_{y, \text{final}} = 0 \) allows us to solve for the height \( h \):\[ h = \frac{v_y^2}{2g} \]This equation tells us the maximum height above the ramp is determined by the initial vertical component of the skier's velocity and the acceleration due to gravity \( g \).
Frictionless Ramp
The scenario assumes a frictionless ramp, where only gravity governs the skier's motion. This simplifies our calculations dramatically because:
  • No energy is lost to heat or frictional forces.
  • All potential energy is converted to kinetic energy.
In real life, friction typically plays a role, reducing the total energy available to be converted into kinetic energy. However, by ignoring friction, we gain insights into ideal system behavior, understanding fundamental principles more clearly.
This assumption is crucial in realizing that mass does not impact the final result, as friction's absence ensures energy conservation purely by gravitational influence.
Energy Transformation
Energy transformation explains the process of changing one form of energy into another. In this skiing exercise, we observe transformations between:
  • Potential Energy (at the start, due to height): \( mgh \)
  • Kinetic Energy (as the skier moves): \( \frac{1}{2}mv^2 \)
The transformation begins as the skier transitions down the ramp. The potential energy diminishes while kinetic energy grows. At the ramp's end, the potential energy is fully converted into kinetic energy.
This cycle of transformation illustrates energy's dynamic conversion in systems, free of resistive forces.
It highlights that the total energy has moved seamlessly from gravitational potential to kinetic, showcasing energy's fundamental conservation nature.

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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?

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