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Chapter 4: Problem 62

(II) A skier moves down a 12\(^\circ\) slope at constant speed.What can you say about the coefficient of friction, \(\mu_k\)? Assume the speed is low enough that air resistance can be ignored.

Short Answer

Expert verified
The coefficient of friction \( \mu_k \) is approximately 0.213.

Step by step solution

01

Analyze the Forces

To find the coefficient of kinetic friction, \( \mu_k \), we need to consider the forces acting on the skier. These include the gravitational force down the slope and the frictional force opposing the motion. Since the skier moves at a constant speed, the net force is zero.
02

Gravitational Force Components

The gravitational force \( F_g \) can be decomposed into two components: one parallel to the slope \( F_{g,\parallel} = mg \sin(\theta) \) and one perpendicular to the slope \( F_{g,\perp} = mg \cos(\theta) \), where \( m \) is the skier's mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the slope angle of \( 12^\circ \).
03

Equilibrium of Forces

In equilibrium, the frictional force \( F_{fr} \) equals the downhill component of the gravitational force: \( F_{fr} = F_{g,\parallel} = mg \sin(\theta) \). The normal force \( N \) equals the perpendicular component: \( N = mg \cos(\theta) \).
04

Expression for Coefficient of Friction

The coefficient of kinetic friction is given by \( \mu_k = \frac{F_{fr}}{N} \). Substituting the expressions from the previous step gives \( \mu_k = \frac{mg \sin(\theta)}{mg \cos(\theta)} = \tan(\theta) \).
05

Calculate the Coefficient

Using \( \theta = 12^\circ \), calculate \( \mu_k = \tan(12^\circ) \). Using a calculator, we find that \( \tan(12^\circ) \approx 0.21256 \).

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Key Concepts

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

Kinetic Friction
When a skier moves down a slope, they experience a force known as kinetic friction. Kinetic friction occurs between two surfaces sliding past each other, one being the skis, and the other, the snow-covered slope. This frictional force opposes the skier's motion. It is quantified by the coefficient of kinetic friction, denoted as \( \mu_k \). This coefficient depends on the nature of both surfaces involved.
  • Friction is a resisting force to motion.
  • Kinetic friction only acts when objects are in relative motion.
If the skier is moving down the slope at a constant speed, it implies that the kinetic friction is exactly balancing the component of gravitational force acting down the slope. This balance is crucial in maintaining a steady speed.
Gravitational Force Components
The gravitational force acting on the skier can be split into two components due to the slope's incline. One component acts parallel to the slope, pulling the skier downhill, and the other acts perpendicular to the slope, affecting the normal force.
  • Parallel component: \( F_{g,\parallel} = mg \sin(\theta) \)
  • Perpendicular component: \( F_{g,\perp} = mg \cos(\theta) \)
Understanding these components is important because they determine how much force contributes to the skier's motion and how much keeps the skier pressed against the slope. The parallel component is what is counteracted by kinetic friction, ensuring a constant speed.
Equilibrium of Forces
In physics, equilibrium means that all the forces acting on an object are balanced. This is the case for a skier moving at constant speed downhill. Here, two main forces need to be considered: the gravitational component acting to push the skier down the slope and the opposing kinetic friction.
  • Kinetic friction: \( F_{fr} = mg \sin(\theta) \)
  • Normal force: \( N = mg \cos(\theta) \)
Since these forces are balanced, the skier doesn't accelerate. This is why we can use the concept of equilibrium to express the coefficient of kinetic friction \( \mu_k \). It ensures that the kinetic friction matches the downhill gravitational pull.
Slope Angle
The slope angle \( \theta \) is crucial in determining the resulting forces acting on the skier. This angle influences how much of the gravitational force contributes to moving the skier down the slope. In our context, the angle is \( 12^\circ \).
  • This angle directly determines the skiing experience, as steeper slopes have higher gravitational force components.
  • The smaller the angle, the more mild the slope, making it easier and slower to ski down.
The tangent of this slope angle mathematically represents the coefficient of kinetic friction for the skier on this particular slope. Specifically, \( \mu_k = \tan(12^\circ) \), allowing us to calculate the frictional coefficient that ensures a constant speed.

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Most popular questions from this chapter

(II) The cable supporting a 2125-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it give the elevator without breaking?

(III) A person jumps from the roof of a house 2.8 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 42 kg, find (\(a\)) his velocity just before his feet strike the ground, and (\(b\)) the average force exerted on his torso by his legs during deceleration.

A 75.0-kg person stands on a scale in an elevator. What does the scale read (in N and in kg) when (\(a\)) the elevator is at rest, (\(b\)) the elevator is climbing at a constant speed of 3.0 m/s, (\(c\)) the elevator is descending at 3.0 m/s, (\(d\)) the elevator is accelerating upward at 3.0 m/s\(^2\), (\(e\)) the elevator is accelerating downward at 3.0 m/s\(^2\)?

Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle to the horizontal, Fig. 4-76 (block \(B\) is above block \(A\)). The masses of the blocks are \(m_A\) and \(m_B\) and the coefficients of friction are \(\mu_A\) and \(\mu_B\). If \(m_A=m_B=\) 5.0 kg and \(\mu_A=\) 0.20 and \(\mu_B=\) 0.30, determine (a) the acceleration of the blocks and (b) the tension in the cord, for an angle \(\theta =\) 32\(^\circ\).

(II) A person pushes a 14.0-kg lawn mower at constant speed with a force of \(F=\) 88.0 N directed along the handle, which is at an angle of 45.0\(^\circ\) to the horizontal (Fig. 4-58). (\(a\)) Draw the free-body diagram showing all forces acting on the mower. Calculate (\(b\)) the horizontal friction force on the mower, then (\(c\)) the normal force exerted vertically upward on the mower by the ground. (\(d\)) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force?
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