/*! 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 84 A 63-kg skier coasts up a snow-c... [FREE SOLUTION] | 91Ó°ÊÓ

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A 63-kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{m} / \mathrm{s}\). After coasting \(1.9 \mathrm{m}\) up the slope, the skier has a speed of \(4.4 \mathrm{m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Short Answer

Expert verified
(a) Work done by friction: -1267.13 J. (b) Kinetic friction force: 667.96 N.

Step by step solution

01

Identify Given Values

Firstly, identify all the values provided in the problem statement: \( m = 63 \, \text{kg} \), initial speed \( v_i = 6.6 \, \text{m/s} \), final speed \( v_f = 4.4 \, \text{m/s} \), distance \( s = 1.9 \, \text{m} \), and the angle \( \theta = 25^{\circ} \).
02

Calculate Initial and Final Kinetic Energy

Compute the initial kinetic energy \( KE_i \) using the formula: \( KE_i = \frac{1}{2} m v_i^2 \). For initial kinetic energy: \[ KE_i = \frac{1}{2} \times 63 \, \text{kg} \times (6.6 \, \text{m/s})^2 \approx 1370.34 \, \text{J} \]Compute the final kinetic energy \( KE_f \) using the formula: \( KE_f = \frac{1}{2} m v_f^2 \). For final kinetic energy: \[ KE_f = \frac{1}{2} \times 63 \, \text{kg} \times (4.4 \, \text{m/s})^2 \approx 608.74 \, \text{J} \]
03

Calculate Change in Kinetic Energy

Find the change in kinetic energy by subtracting the final kinetic energy from the initial kinetic energy:\[ \Delta KE = KE_f - KE_i = 608.74 \, \text{J} - 1370.34 \, \text{J} = -761.6 \, \text{J} \]
04

Calculate Work Done Against Gravity

Next, calculate the work done against gravity, which is given by:\[ W_g = m g s \sin(\theta) \]Where \( g \) is the acceleration due to gravity \( \approx 9.81 \, \text{m/s}^2 \), giving:\[ W_g = 63 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.9 \, \text{m} \times \sin(25^{\circ}) \approx 505.53 \, \text{J} \]
05

Use Work-Energy Theorem

Apply the work-energy theorem which states that the total work done is equal to the change in kinetic energy. Therefore,\[ W_f + W_g = \Delta KE \]Since \( W_g \) is known, rearrange to solve for \( W_f \):\[ W_f = \Delta KE - W_g = -761.6 \, \text{J} - 505.53 \, \text{J} = -1267.13 \, \text{J} \]The negative sign indicates work done by the frictional force is opposing the motion.
06

Calculate Kinetic Frictional Force

Now find the frictional force using:\[ W_f = f_k \cdot s \]Solve for \( f_k \):\[ f_k = \frac{W_f}{s} = \frac{-1267.13 \, \text{J}}{1.9 \, \text{m}} \approx -667.96 \, \text{N} \]The magnitude of the force is \( 667.96 \, \text{N} \), indicating the direction is opposite to motion.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. The more massive or faster the object moves, the more kinetic energy it carries. This is crucial in physics as it helps in explaining how and why objects move the way they do. Kinetic energy (\( KE \)) is computed using the equation:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity.
To fully comprehend kinetic energy, consider a skier sliding down a hill. As gravity pulls the skier downward, it gains speed and thus kinetic energy increases.
From the exercise, when the skier's speed changes from \( 6.6 \, \text{m/s} \) to \( 4.4 \, \text{m/s} \) during the ascent, their kinetic energy decreases, reflecting a transfer or conversion of energy.
Work-Energy Theorem
The Work-Energy Theorem is pivotal in understanding energy changes. It states that the work done on an object is equal to the change in its kinetic energy. This can be expressed with the equation:
  • \( W = \Delta KE \)
where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy.
In the exercise, the skier’s kinetic energy changes due to both the work done by friction and the work done against gravity.
This theorem outlines the balance of forces doing work on the skier: gravity helps propel them, while friction works against this motion.
The difference between initial and final kinetic energy gives an insight into how much work was required to overcome opposing forces such as friction in this system.
Frictional Force
Frictional force plays an essential role in most real-life physics problems, including this one. It is the force that opposes the motion of an object. When a skier moves up or down an inclined plane, friction is always acting opposite to the direction of movement.
Calculated using the relationship:
  • \( W_f = f_k \cdot s \)
where \( W_f \) is the work done by friction and \( s \) is the distance traveled.
From the exercise, the frictional work is negative, indicating resistance to motion.
The negative work done by friction lowered the skier's kinetic energy, slowing them down.
It is simply a force that needs to be overcome when trying to move objects across surfaces, affecting their speed and energy balance.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, often used in physics to analyze forces acting along a slope. It simplifies the study of forces and motion. Whenever objects move up or down these planes, gravity has a component acting along the plane.
Breaking down gravity into components helps in understanding how other forces, such as normal force and friction, come into play.
  • The gravitational force acting parallel to the plane is \( m g \sin(\theta) \).
This component is crucial as it dictates how easily an object like a skier can move up or down the slope.
In this exercise, understanding the inclined plane helps calculate how gravity assists or opposes the skier’s motion.
The steeper the incline, the more significant the gravitational pull along the plane, thereby affecting the skier's kinetic energy and the overall work done during the motion.
Studying inclined planes is fundamental not just in understanding theoretical concepts but also in practical applications in various fields.

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Most popular questions from this chapter

In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of \(140 \mathrm{m}\). The average mass of each skier is \(65 \mathrm{kg}\). What is the average power provided by the tension in the cable pulling the lift?

A 2.00-kg rock is released from rest at a height of 20.0 m. Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: \(20.0,10.0,\) and \(0 \mathrm{m}\)

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta .\) One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0}\). On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

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?

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water \(5.00 \mathrm{m}\) from the end of the slide in a time of \(0.500 \mathrm{s}\) after leaving the slide. Ignoring friction and air resistance, find the height \(H\) in the drawing.

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