/*! 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 62 A \(60.0-\mathrm{kg}\) skier sta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 62

A \(60.0-\mathrm{kg}\) skier starts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathbf{k J}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and Penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Short Answer

Expert verified
(a) 30.38 m/s; (b) 12.75 m/s; (c) 1950.25 N.

Step by step solution

01

Determine Potential Energy at the Top

To find the skier's potential energy at the top of the slope, calculate using the formula for gravitational potential energy: \[ PE = m \cdot g \cdot h \]where \( m = 60.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 65.0 \, \text{m} \). Substituting these values, we get:\[ PE = 60.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 65.0 \, \text{m} = 38223 \, \text{J} \] or \( 38.2 \, \text{kJ} \).
02

Calculate Work Done by Friction

The frictional force does \(-10.5 \, \text{kJ}\) of work, which converts into a reduction in mechanical energy as the skier descends.
03

Apply Conservation of Energy to Find Speed at Bottom

The energy conversion can be calculated as follows:\[ KE_{bottom} = PE_{top} + W_{frictional} \]Where \( KE_{bottom} \) is the kinetic energy at the bottom of the slope and is given by:\[ KE_{bottom} = \frac{1}{2} m v^2 \]Thus:\[ \frac{1}{2} \times 60.0 \, \text{kg} \times v^2 = 38.2 \, \text{kJ} - 10.5 \, \text{kJ} \]\[ 30 v^2 = 27.7 \, \text{kJ} \]\[ v^2 = \frac{27700 \, \text{J}}{30} \approx 923.33 \]\[ v = \sqrt{923.33} \approx 30.38 \, \text{m/s} \]
04

Calculate Work Done Across the Snow Patch

Here, calculate the forces acting on the skier as she crosses the patch:- Friction force \( F_{fric} = \mu_k \times m \times g = 0.20 \times 60.0 \times 9.81 = 117.72 \, \text{N} \).- Air resistance \( F_{air} = 160 \, \text{N} \).The total work done against these forces is:\[ W = (F_{fric} + F_{air}) \times d = (117.72 + 160) \times 82.0 = 22842.64 \, \text{J} \].
05

Determine Final Velocity after Snow Patch

The work-energy principle tells us:\[ \Delta KE = W \]Thus:\[ \frac{1}{2} \times 60.0 \times (v_{after}^2 - v_{before}^2) = -22842.64 \]\[ 30 \times (v_{after}^2 - 923.33) = -22842.64 \]\[ v_{after}^2 = 923.33 - \frac{22842.64}{30} = 162.57 \]\[ v_{after} = \sqrt{162.57} \approx 12.75 \, \text{m/s} \]
06

Determine Force Exerted by the Snowdrift

Since the skier comes to a stop after penetrating the snowdrift, the kinetic energy is completely converted into work done by the snowdrift:\[ \Delta KE = -W \]\[ \frac{1}{2} \times 60.0 \times 12.75^2 = F_{snow} \times 2.5 \]\[ 4875.625 = F_{snow} \times 2.5 \]\[ F_{snow} = \frac{4875.625}{2.5} \approx 1950.25 \, \text{N} \]

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 91影视!

Key Concepts

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

Work-Energy Principle
The Work-Energy Principle is like a bridge connecting the concepts of work and energy. It explains how the work done on an object affects its energy. In mechanics, this principle is incredibly useful to calculate changes in kinetic energy and potential energy. Work done on an object can change its energy, and this principle can be mathematically expressed as:
\[ \Delta KE = W \]
This means that the change in kinetic energy \( \Delta KE \) of a system is equal to the work \( W \) done on that system. This principle helps us understand various scenarios, such as when a skier moves down a ski slope. As the skier descends, both gravitational work and frictional work act upon the skier, affecting the total mechanical energy of the system.
  • If work is done on the system, like friction that does negative work, it reduces the system's energy.
  • Conversely, positive work adds energy to the system.
In practical applications, such as the skier example, the Work-Energy Principle allows us to calculate the skier's velocity at various points, after accounting for all forces doing work on her.
Kinetic Energy
Kinetic Energy is the energy that an object possesses due to its motion. When discussing kinetic energy in mechanics, it is usually represented by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Here, \( KE \) stands for Kinetic Energy, \( m \) is the mass of the object, and \( v \) is its velocity. Kinetic energy depends on both the mass and the velocity of the object, meaning that more massive objects or those moving quicker have greater kinetic energy.
Consider the scenario of a skier descending a slope. Initially, the skier starts from rest and has zero kinetic energy. As she moves downhill, her velocity increases due to gravity, and consequently, her kinetic energy also increases. However, factors such as friction can reduce this energy by doing work against the skier, slowing her down.
  • At the bottom of the slope, her kinetic energy is influenced by the initial potential energy and the work done by friction.
  • Kinetic energy variations allow us to calculate her speed, using the Work-Energy Principle.
Understanding kinetic energy transformations is essential in predicting the motion of objects across various situations in mechanics.
Potential Energy
Potential Energy is the stored energy in an object due to its position or state. In the context of mechanics, gravitational potential energy is one of the most common forms, given by the expression:
\[ PE = mgh \]
where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is height. This formula calculates the energy stored in an object raised above ground level. This type of energy is stored due to gravitational forces acting on the object.
In our skier example, the skier begins with a certain amount of gravitational potential energy at the top of the slope. This potential energy depends on her mass, the height of the slope, and gravity. As she descends, this stored energy converts into kinetic energy, causing her to accelerate.
  • At the top of the slope, potential energy is at its maximum because of the great height.
  • As she moves down, potential energy decreases, converting into kinetic energy.
Understanding potential energy helps predict how energy conversion provides motion or change in a system. Conservation laws tell us that the total mechanical energy remains constant if no non-conservative forces, like friction, do work.

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

A cutting tool under microprocessor control has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{j}},\) a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{F}\) if this displacement is along the straight line \(y=x\) that connects, these two points. (b) Calculate the work done on the tool by \(\vec{F}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}\) , \(y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{F}\) along these two paths. Is \(\vec{F}\) conservative or nonconservative? Explain.

You are asked to design a spring that will give a \(1160-\mathrm{kg}\) satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g}\) . The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

A stone of mass \(m\) is thrown upward at an angle \(\theta\) above the horizontal and feels no appreciable air resistance. Use conservation of energy to show that at its highest point, it is a distance \(v_{0}^{2}\left(\sin ^{2} \theta\right) / 2 g\) above the point where it was launched. (Hint: \(v_{0}^{2}=v_{\mathrm{ax}}^{2}+v_{\mathrm{oy}}^{2} . )\)

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Gravity in Two Dimensions. Two point masses, \(m_{1}\) and \(m_{2},\) lie in the \(x y\) -plane, with \(m_{1}\) held in place at the origin and \(m_{2}\) free to move a distance \(r\) away at a point \(P\) having coordinates \(x\) and \(y\) (Fig. 7.27\()\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant. (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}\right)^{3 / 2}} \quad \text { and } \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x\) and \(y . )(\text { b) Show that the magnitude}\) of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\) (c) Does \(m_{1}\) attract or repel \(m_{2} ?\) How do you know?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Translational Dynamics

Read Explanation

Engineering Physics

Read Explanation

Fields in Physics

Read Explanation

Electromagnetism

Read Explanation

Thermodynamics

Read Explanation
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.