/*! 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 52 Ski Jump Ramp. You are designing... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Ski Jump Ramp. You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height \(h\) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 \(\mathrm{m} / \mathrm{s}\) as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 \(\mathrm{m} / \mathrm{s}\) when they reach the bottom of the ramp. You determine that for a \(85.0-\mathrm{kg}\) skier with good form, friction and air resistance will do total work of magnitude 4000 \(\mathrm{J}\) on him during his run down the slope. What is the maximum height \(h\) for which the maximum safe speed will not be exceeded?

Short Answer

Expert verified
The maximum height is approximately 50.5 meters.

Step by step solution

01

Write down the given data

Initial speed of the skier, \(v_i = 2.0 \, \text{m/s}\). Final speed at the bottom, \(v_f = 30.0 \, \text{m/s}\). Mass of the skier, \(m = 85.0 \, \text{kg}\). Work done against friction and air resistance, \(W = 4000 \, \text{J}\). We need to find the vertical height \(h\).
02

Apply conservation of energy principle

The mechanical energy at the starting point plus any work done by non-conservative forces (like friction) should equal the mechanical energy at the bottom of the ramp. The equation is:\[KE_i + PE_i + W = KE_f + PE_f\]Where \(KE_i\) is the initial kinetic energy, \(PE_i\) is the initial potential energy, \(KE_f\) is the final kinetic energy, and \(PE_f\) is the final potential energy.
03

Express kinetic and potential energies

The initial kinetic energy \(KE_i = \frac{1}{2}mv_i^2\), potential energy at the top \(PE_i = mgh\), final kinetic energy \(KE_f = \frac{1}{2}mv_f^2\), and potential energy at the bottom \(PE_f = 0\) since all potential energy has been converted to kinetic energy or lost to work. Substitute and simplify:\[\frac{1}{2}mv_i^2 + mgh - W = \frac{1}{2}mv_f^2\]
04

Solve for the height h

Substitute the given values into the equation:\[\frac{1}{2}(85)(2.0)^2 + 85 \cdot 9.8 \cdot h - 4000 = \frac{1}{2}(85)(30.0)^2\]Simplifying this:\[0.5(85)(4) + 85(9.8)h - 4000 = 0.5(85)(900)\]\[170 + 833h - 4000 = 38250\]\[833h = 38250 - 170 + 4000\]\[833h = 42080\]\[h = \frac{42080}{833} \approx 50.5 \, \text{m}\]
05

State the result

The maximum height \(h\) for which the skiers can safely reach the bottom of the ramp without exceeding the speed limit is approximately 50.5 meters.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 possesses due to its motion. It depends on two main factors: the mass of the object and its velocity. In mathematical terms, kinetic energy (\( KE \)) can be expressed as:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass and \( v \) is the velocity. For a moving skier, kinetic energy changes as they move down the slope reflecting the skier's speed changes.
At the beginning of a ski jump, the skier has an initial kinetic energy derived from their initial push-off speed. This initial kinetic energy at the gate can be calculated using the skier's mass and their initial speed of 2.0 \( \text{m/s} \). As the skier approaches the bottom with a higher speed (up to 30.0 \( \text{m/s} \)), their kinetic energy increases.
Ultimately, kinetic energy helps us understand how fast the skier will be moving at any given point on the jump on the ramp.
Potential Energy
Potential energy is the stored energy of an object due to its position. In the context of ski jumps, it's particularly associated with gravitational potential energy. This energy is determined by the height of the skier above a reference point, such as the bottom of the ramp. Mathematically, potential energy (\( PE \)) is expressed as:
  • \( PE = mgh \)
where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height.
For the skier at the top of the ramp, this energy is at its maximum. As they descend, potential energy is converted into kinetic energy.
This conversion helps achieve the maximum speed at the bottom of the ramp, showcasing the fundamental energy transformations occurring during the ski jump.
Work-Energy Principle
The work-energy principle is essential in analyzing the movement of skiers in a jump. It states that the work done by the forces acting on an object is equal to the change in the object's kinetic energy. It's articulated as:
  • Work done + Initial Mechanical Energy = Final Mechanical Energy
In the context of the ski jump, non-conservative forces such as friction and air resistance do work on the skier, altering their energy state.
In this exercise, 4000 \( \text{J} \) of energy is spent overcoming these forces. The initial mechanical energy (sum of initial kinetic and potential energy) decreases by this amount, influencing the skier's speed as they reach the bottom.
By understanding this principle, we can calculate how much potential energy needs to be present at the top to account for energy lost due to work done by friction and maintain safety speed limits.
Ski Jump Physics
Ski jump physics combines various principles of mechanics to ensure skiers safely navigate the ramp and achieve optimal performance. Central to this is energy conservation, where potential energy at the top of the ramp becomes kinetic energy at the bottom. The skier's design speed is carefully calculated to ensure safe descent.
The maximum speed attainable at the bottom of the ramp must not exceed 30.0 \( \text{m/s} \) for safety. This involves a delicate balance between initial kinetic energy, potential energy at the top, and energy work done against friction.
Calculating the ramp's height is crucial as it determines the initial potential energy available for conversion. By precisely estimating this height, we ensure the ski jump works within physics laws, making ski jumping both exciting and safe.
Overall, ski jump physics is a practical application of energy principles, demonstrating how conservation laws guide sports safety and performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the middle of the gym floor, as shown in Fig. \(7.26 .\) You take your physics book and push it from one person to the other. The book has a mass of 1.5 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the book and the floor is \(\mu_{\mathrm{x}}=0.25 .\) (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

When it is burned, 1 gallon of gasoline produces \(1.3 \times 10^{8} \mathrm{J}\) of energy. A \(1500-\mathrm{kg}\) car accelerates from rest to 37 \(\mathrm{m} / \mathrm{s}\) in 10 \(\mathrm{s}\) . The engine of this car is only 15\(\%\) efficient (which is typical), meaning that only 15\(\%\) of the energy from the combustion of the gasoline is used to accelerate the car. The rest goes into things like the internal kinetic energy of the engine parts as well as heating of the exhaust air and engine. (a) How many gallons of gasoline does this car use during the acceleration? (b) How many such accelerations will it take to burn up 1 gallon of gas?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 \(\mathrm{kg}\) clings to one end of the stick, and a mouse with mass 0.200 \(\mathrm{kg}\) clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, whit are their speeds as the rod swings through a vertical position?

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He starts from rest. He moves as fast at the bottom as if be had stepped off a platform a distance \(h \leq d\) above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special cases of \(h=d\) and \(h=0 ?\) (b) Find a numerical value for the average friction force a \(75-\mathrm{kg}\) fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m},(\mathrm{c})\) In terms of \(g, h,\) and \(d\) , what is the speed of the fireman when he is a distance y above the bottom of the pole?

An object has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (See Problem \(6.96 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2} .\) The object moves along the following path: (1) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\overrightarrow{\boldsymbol{F}}\) for each leg of the path and for the complete round trip. (c) Is \(\overrightarrow{\boldsymbol{F}}\) conservative or nonconservative? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.