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Chapter 7: Problem 83

A \(68 \mathrm{~kg}\) skier approaches the foot of a hill with a speed of \(15 \mathrm{~m} / \mathrm{s}\). The surface of this hill slopes up at \(40.0^{\circ}\) above the horizontal and has coefficients of static and kinetic friction of 0.75 and \(0.25,\) respectively, with the skis. (a) Use energy conservation to find the maximum height above the foot of the hill that the skier will reach. (b) Will the skier remain at rest once she stops, or will she begin to slide down the hill? Prove your answer.

Short Answer

Expert verified
(a) Use energy conservation to find height. (b) Compare friction forces to determine if skier slides.

Step by step solution

01

Understand the Problem

To find how high the skier climbs, we'll use energy conservation. We know the skier's initial kinetic energy and must find how far up the hill they go before stopping. We also must check if friction can hold them on the hill.
02

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the skier is given by: \[KE = \frac{1}{2} m v^2\]Substitute: \[m = 68 \, \text{kg}, \ v = 15 \, \text{m/s}\]Thus, \[ KE = \frac{1}{2} \times 68 \times 15^2 = 7650 \, \text{J}\]
03

Determine the Work Done by Friction

The frictional force (f) causes a work done given as: \[W_{\text{fric}} = f \times d\]First, calculate the normal force and the frictional force. For a slope, \[f = \mu_k N = \mu_k mg \cos \theta\]Using \( \mu_k = 0.25 \), \( m = 68 \, \text{kg} \), and \( \theta = 40^\circ \), \[N = mg \cos \theta = 68 \times 9.8 \times \cos 40^\circ \]Calculate \( f \) and then determine \( W_{\text{fric}} \) as part of the work-energy principle.
04

Apply Energy Conservation

The concept of energy conservation states: \[KE_{\text{initial}} - W_{\text{fric}} = PE_{\text{max}}\]The potential energy at height \( h \) is given by: \[PE = mgh \]Set \( mgh = KE_{\text{initial}} - W_{\text{fric}} \) and solve for \( h \). Use the calculated \( W_{\text{fric}} \) from the previous step.
05

Determine if Skier Will Slide Again

To check if the skier will remain at rest, compare static friction available with the force component along the slope. \[F_{\text{along}} = mg \sin \theta\] Static friction \( f_{\text{static}} = \mu_s mg \cos \theta \) For the skier to remain at rest: \[f_{\text{static}} \geq F_{\text{along}}\] Determine which one is greater to conclude if the skier slides.

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Key Concepts

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

Kinetic Energy
When the skier approaches the hill, she has a certain amount of kinetic energy due to her motion. This energy can be calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the skier and \( v \) is her velocity. In this case, \( m = 68 \, \text{kg} \) and \( v = 15 \, \text{m/s} \), resulting in a kinetic energy of 7650 joules. Kinetic energy is an indicator of how much work the skier can do before coming to a stop, such as climbing up a hill. It transforms into potential energy as the skier ascends.
Potential Energy
As the skier moves up the slope, her kinetic energy is converted into potential energy. Potential energy is the energy stored in an object due to its height above the ground. It can be calculated using:\[ PE = mgh \]where \( m \) is mass, \( g \) is the acceleration due to gravity (9.8 \( \text{m/s}^2 \)), and \( h \) is the height the skier climbs. When the skier reaches the maximum height, all the initial kinetic energy is converted into potential energy, minus any work done against friction.
Friction
Friction plays a crucial role in determining how far the skier can travel up the hill. It is a force that opposes motion between two surfaces that are in contact. In this scenario, the friction is between the skier's skis and the snowy slope. Friction reduces the skier's kinetic energy, slowing her down as she climbs. The work done by friction (\( W_{\text{fric}} \)) is subtracted from the total energy available for climbing, as frictional forces convert some of this energy into heat.
Static Friction
Once the skier reaches the top and stops, static friction is the force that will try to keep her stationary on the slope. Static friction is greater than kinetic friction and is calculated using:\[ f_{\text{static}} = \mu_s mg \cos \theta \]where \( \mu_s \) is the coefficient of static friction. In this exercise, the static friction must be enough to counteract the component of gravitational force pulling the skier back down the slope. If the static friction exceeds this force, the skier stays at rest. Otherwise, she slides down.
Kinetic Friction
Kinetic friction comes into play if the skier begins to move again after stopping. It is the frictional force experienced by an object already in motion and is usually less than static friction. Kinetic friction is calculated using:\[ f = \mu_k mg \cos \theta \]where \( \mu_k \) is the coefficient of kinetic friction. For the skier, kinetic friction is responsible for reducing her speed as she moves up or down the slope. It acts continuously to convert part of her mechanical energy into heat, slowing her progress and affecting how far she travels on the hill.

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Most popular questions from this chapter

A tennis player hits a \(58.0 \mathrm{~g}\) tennis ball so that it goes straight up and reaches a maximum height of \(6.17 \mathrm{~m} .\) How much work does gravity do on the ball on the way up? On the way down?

During the calibration process, the cantilever is observed to deflect by \(0.10 \mathrm{nm}\) when a force of \(3.0 \mathrm{pN}\) is applied to it. What deflection of the cantilever corresponds to a force of \(6.0 \mathrm{pN} ?\) A. \(0.07 \mathrm{nm}\) C. \(0.20 \mathrm{nm}\) B. \(0.14 \mathrm{nm}\) D. \(0.40 \mathrm{nm}\)

A bullet is fired into a large stationary absorber and comes to rest. Temperature measurements of the absorber show that the bullet lost \(1960 \mathrm{~J}\) of kinetic energy, and high-speed photos of the bullet show that it was moving at \(965 \mathrm{~m} / \mathrm{s}\) just as it struck the absorber. What is the mass of the bullet?

A spring is \(17.0 \mathrm{~cm}\) long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to \(25.0 \mathrm{~N}\), causing the spring to stretch to a length of \(19.2 \mathrm{~cm}\). (a) What is the force constant of this spring? (b) How much work was required to stretch the spring from \(17.0 \mathrm{~cm}\) to \(19.2 \mathrm{~cm} ?\) (c) How long will the spring be if the \(25 \mathrm{~N}\) force is replaced by a \(50 \mathrm{~N}\) force?

A \(61 \mathrm{~kg}\) skier on level snow coasts \(184 \mathrm{~m}\) to a stop from a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (a) Use the work-energy theorem to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a \(75 \mathrm{~kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.
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