/*! 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 92 A Sledding Contest. You are in a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Sledding Contest. You are in a sledding contest where you start at a height of \(40.0 \mathrm{m}\) above the bottom of a valley and slide down a hill that makes an angle of \(25.0^{\circ}\) with respect to the horizontal. When you reach the valley, you immediately climb a second hill that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The winner of the contest will be the contestant who travels the greatest distance up the second hill. You must now choose between using your flat-bottomed plastic sled, or your "Blade Runner," which glides on two steel rails. The hill you will ride down is covered with loose snow. However, the hill you will climb on the other side is a popular sledding hill, and is packed hard and is slick. The two sleds perform very differently on the two surfaces, the plastic one performing better on loose snow, and the Blade Runner doing better on hard-packed snow or ice. The performances of each sled can be quantified in terms of their respective coefficients of kinetic friction on the two surfaces. For the plastic sled: \(\mu=0.17\) on loose snow, and \(\mu=0.15\) on packed snow or ice. For the Blade Runner, \(\mu=0.19\) on loose snow, and \(\mu=0.07\) on packed snow or ice. Assuming the two hills are shaped like inclined planes, and neglecting air resistance, (a) how far does each sled make it up the second hill before stopping? (b) Assuming the total mass of the sled plus rider is \(55.0 \mathrm{kg}\) in both cases, how much work is done by nonconservative forces (over the total trip) in each case?

Short Answer

Expert verified
Blade Runner travels further up the second hill. Nonconservative work is higher for the plastic sled.

Step by step solution

01

Calculating initial velocity at the bottom of the hill

To solve part (a), we first calculate the speed at the bottom of the valley for each sled using energy conservation. The potential energy at the top is converted to kinetic energy at the bottom minus the work done by friction on the first hill.For the plastic sled:1. Gravitational potential energy at the top: \[ PE = mgh = 55.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 40.0 \, \mathrm{m}= 21560 \, \mathrm{J} \]2. Work done by friction on the first hill: \[ W_{f1} = \mu mg \cos(\theta_1) d_1 = 0.17 \times 55.0 \times 9.8 \times \cos(25^\circ) \times \frac{40.0}{\sin(25^\circ)} \]3. Solve for speed at the bottom, \(v_1\): \[ \frac{1}{2}mv_1^2 = PE - W_{f1} \quad \Rightarrow v_1 = \sqrt{\frac{2 \cdot (PE - W_{f1})}{m}} \] Repeat these calculations for the Blade Runner with its respective \(\mu\).
02

Calculating distance up the second hill

For both sleds, calculate the distance they reach on the second hill. The kinetic energy at the bottom is converted into potential energy at the height they reach minus the work done by friction.For the plastic sled:1. Initial kinetic energy at the bottom:\[ KE = \frac{1}{2}mv_1^2 \]2. Final potential energy at height h:\[ PE = mgh_2 \]3. Work done by friction on the second hill:\[ W_{f2} = \mu mg \cos(\theta_2) d_2 \]4. Balance of energies: \[ \frac{1}{2}mv_1^2 = mgh_2 + W_{f2} \]Since \(h_2 = d_2 \sin(15^\circ)\), solve for \(d_2\) using \[ \frac{1}{2}mv_1^2 = mgd_2 \sin(15^\circ) + \mu mg \cos(15^\circ) d_2 \]. Do the same for the Blade Runner.
03

Calculating work done by nonconservative forces

Calculate the work done by nonconservative forces for each sled, which is the total work done by friction on both hills.For the plastic sled:1. Work on the first hill : \( W_{f1} = \mu_1 mg \cos(\theta_1) d_1 \)2. Work on the second hill: \( W_{f2} = \mu_2 mg \cos(\theta_2) d_2 \)3. Total work by nonconservative forces: \( W_{nc} = W_{f1} + W_{f2} \)Repeat these calculations for the Blade Runner using \(\mu\) values respective to each sled on each hill.

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 a moving object possesses due to its motion. When you're sledding down a hill, the sled's kinetic energy increases as it gains speed. This energy is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. To illustrate, when you slide down the hill from a height, the potential energy due to gravity is converted into kinetic energy, increasing your sled's speed.
However, it's important to remember that not all potential energy becomes kinetic energy due to the friction between the sled and the snow. The work done against friction reduces the kinetic energy achieved. This means that even though the sled starts with a set amount of potential energy, less is available to convert to kinetic energy because some is "lost" through friction.
Potential Energy
Potential energy relates to an object's position and is stored energy that can be converted into kinetic energy. When a sled starts at the top of a hill, it has a lot of gravitational potential energy due to its height. The formula for gravitational potential energy is \( PE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( h \) is the height. As the sled descends, this potential energy begins to decrease, transforming into kinetic energy, causing the sled to gain speed. However, because friction acts on the sled, not all energy is converted, and some remains as potential energy at various sled heights or is used overcoming friction.
Friction Coefficient
The friction coefficient, denoted as \( \mu \), measures how much frictional force acts between two surfaces. Different sled designs and surfaces (like loose snow vs. packed snow) will have various coefficients of friction, affecting how far the sled can slide. In a sledding context:
  • The plastic sled has coefficients of 0.17 on loose snow and 0.15 on packed snow.
  • The Blade Runner sled has coefficients of 0.19 on loose snow and 0.07 on packed snow.
A lower friction coefficient means less resistance, allowing an object to travel further. Therefore, the Blade Runner performs better on packed snow due to its lower friction coefficient, allowing it to ascend further up the second hill when compared to the plastic sled.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. Sledding down an inclined plane allows for the conversion of potential energy to kinetic energy, and it impacts the level of friction the sled will experience.

In our sledding contest, the angles of the hills influence the direction of gravitational force and friction. The steepness of the first hill (25°) allows a fast acceleration because gravity pulls the sled downward more directly compared to the less steep second hill (15°). When climbing the second hill, the angle also affects how high each sled can go before friction and gravity stops it.

Energy Conservation
Energy conservation is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form to another. In sledding, the energy transitions from potential to kinetic and back to potential as the sled travels up and down hills. However, energy is not always fully conserved in the form we might expect because some is spent overcoming friction.
  • At the start, the sled's potential energy is at its peak.
  • As it descends, potential energy is converted into kinetic energy.
  • Climbing the next hill, kinetic energy again converts back into potential energy.
  • Throughout, friction "consumes" some energy, limiting the distance the sled can travel up after descending.
Understanding this chain of transformations helps in calculating how friction influences the performance of different sleds on various surfaces.

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

A55-kg box is being pushed a distance of \(7.0 \mathrm{m}\) across the floor by a force \(\overrightarrow{\mathbf{P}}\) whose magnitude is \(160 \mathrm{N}\). The force \(\overrightarrow{\mathbf{P}}\) is parallel to the displacement of the box. The coefficient of kinetic friction is \(0.25 .\) Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of \(14.6^{\circ}\) with the horizontal. What is the change in the skier's gravitational potential energy?

The cheetah is one of the fastest-accelerating animals, because it can go from rest to \(27 \mathrm{m} / \mathrm{s}\) (about \(60 \mathrm{mi} / \mathrm{h}\) ) in \(4.0 \mathrm{s}\). If its mass is \(110 \mathrm{kg}\), determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{N}\). What is the magnitude of the pulling force exerted by his wife?

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{m} / \mathrm{s}\) in a time of \(3.5 \mathrm{s}\) During this time it climbs to a height of \(8.2 \mathrm{m} .\) What is the average power generated by the lifting force?

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.