/*! 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 101 A skier is pulled up a slope at ... [FREE SOLUTION] | 91Ó°ÊÓ

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A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at \(25.0^{\circ}\) with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is \(55.0 \mathrm{~kg},\) and the coefficient of kinetic friction between the skis and the snow is \(0.120 .\) Find the magnitude of the force that the tow bar exerts on the skier

Short Answer

Expert verified
The force exerted by the tow bar is 237 N.

Step by step solution

01

Identify Forces Acting on the Skier

The skier is moving at a constant velocity up the slope, which means the net force is zero. The forces acting on the skier include the gravitational force, the normal force, the frictional force, and the force from the tow bar. The gravitational force can be split into two components: one parallel to the slope and one perpendicular to the slope.
02

Calculate Gravitational Force Components

The gravitational force (\( F_g \)) can be calculated as: \[ F_g = mg \]where \( m = 55.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).The parallel component \( F_{g_{ ext{parallel}}} \) is:\[ F_{g_{ ext{parallel}}} = F_g \sin\theta \]The perpendicular component \( F_{g_{ ext{perpendicular}}} \) is:\[ F_{g_{ ext{perpendicular}}} = F_g \cos\theta \]Substitute \( \theta = 25.0^{\circ} \).
03

Calculate Normal Force

The normal force (\( F_N \)) is equal to the perpendicular component of the gravitational force when there is no vertical acceleration:\[ F_N = F_{g_{ ext{perpendicular}}} = mg \cos\theta \]Substitute the known values to calculate the normal force.
04

Calculate Frictional Force

The frictional force (\( F_f \)) can be calculated using the coefficient of kinetic friction (\( \mu_k \)):\[ F_f = \mu_k F_N \]Substitute \( \mu_k = 0.120 \) and the value of the normal force calculated in the previous step.
05

Apply Newton's First Law to Find Tow Bar Force

Using Newton's first law, since the skier is moving at a constant velocity, the net force parallel to the slope is zero. Therefore, the force from the tow bar (\( F_t \)) must balance the sum of the frictional force and the gravitational component parallel to the slope:\[ F_t = F_f + F_{g_{ ext{parallel}}} \]Substitute the values calculated previously to solve for \( F_t \).

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Key Concepts

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

Understanding Kinetic Friction
Kinetic friction is a crucial concept, especially when studying objects moving along surfaces like our skier on a slope. When an object slides across a surface, kinetic friction opposes its movement. It plays a vital role in determining how much force is needed to maintain that movement at a constant speed.
Kinetic friction can be calculated with the formula:
  • \( F_f = \mu_k F_N \).
Here, \( F_f \) is the force of friction, \( \mu_k \) is the coefficient of kinetic friction, and \( F_N \) is the normal force.
In this scenario, the coefficient of kinetic friction is given as 0.120. This means that just 12% of the normal force acts in opposition to the skier's motion. Understanding these values helps us see how friction directly affects the force required by the tow bar to keep the skier moving uphill at a constant speed.
Gravitational Force Components
Gravitational force is an essential factor when dealing with inclined planes. It always acts downward towards the center of the Earth. On an incline, this force can be split into two components:
- **Parallel to the slope:** This component, represented as \( F_{g_{\text{parallel}}} \), pulls the object down the slope. It can be calculated as:
  • \( F_{g_{\text{parallel}}} = mg \sin \theta \).
- **Perpendicular to the slope:** This component, noted as \( F_{g_{\text{perpendicular}}} \), acts perpendicular to the slope’s surface and is calculated by:
  • \( F_{g_{\text{perpendicular}}} = mg \cos \theta \).
These components help in understanding how the gravitational force affects objects on slopes. In our exercise, \( F_{g_{\text{parallel}}} \) is one of the forces balanced by the tow bar to ensure the skier moves up the slope without acceleration.
Newton's First Law in Action
Newton's First Law, often called the law of inertia, tells us that an object will stay at rest or move at a constant velocity unless acted on by a net external force.
For the skier, this means moving at the same velocity requires that the total forces parallel to the slope equal zero. Thus, the tow bar exerts a force equal to the combined effects of the parallel gravitational force and the kinetic friction.
This principle implies:
  • Force from the tow bar (\( F_t \)) counteracts both kinetic friction and the force pulling the skier down the slope (gravitational force component parallel to the slope).
By applying Newton's First Law, we find that:
  • \( F_t = F_f + F_{g_{\text{parallel}}} \).
Thus, this law helps us compute the particular force needed by the tow bar to keep the skier gliding smoothly up the incline.

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Most popular questions from this chapter

A large rock and a small pebble are held at the same height above the ground. (a) Is the gravitational force exerted on the rock greater than, less than, or equal to that exerted on the pebble? Justify your answer. (b) When the rock and the pebble are released, is the downward acceleration of the rock greater than, less than, or equal to that of the pebble? Why? Problem A \(5.0-\mathrm{kg}\) rock and a \(3.0 \times 10^{-4} \mathrm{~kg}\) pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the acceleration of each object when released. Verify that your answers are consistent with your answers to the Concept Questions.

The alarm at a fire station rings and an \(86-\mathrm{kg}\) fireman, starting from rest, slides down a pole to the floor below (a distance of \(4.0 \mathrm{~m}\) ). Just before landing, his speed is \(1.4 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

A person, sunbathing on a warm day, is lying horizontally on the deck of a boat. Her mass is \(59 \mathrm{~kg},\) and the coefficient of static friction between the deck and her is \(0.70 .\) Assume that the person is moving horizontally, and that the static frictional force is the only force acting on her in this direction. (a) What is the magnitude of the static frictional force when the boat moves with a constant velocity of \(+8.0 \mathrm{~m} / \mathrm{s} ?\) (b) The boat speeds up with an acceleration of \(1.6 \mathrm{~m} / \mathrm{s}^{2},\) and she does not slip with respect to the deck. What is the magnitude of the static frictional force that acts on her? (c) What is the magnitude of the maximum acceleration the boat can have before she begins to slip relative to the deck?

A damp washcloth is hung over the edge of a table to dry. Thus, part (mass \(=m_{\text {on }}\) ) of the washcloth rests on the table and part (mass \(=m_{\text {off }}\) ) does not. The coefficient of static friction between the table and the washcloth is \(0.40 .\) Determine the maximum fraction \(\left[m_{\text {off }} /\left(m_{\text {on }}+m_{\text {off }}\right)\right]\) that can hang over the edge without causing the whole washcloth to slide off the table

A rock of mass 45 kg accidentally breaks loose from the edge of a cliff and falls straight down. The magnitude of the air resistance that opposes its downward motion is \(18 \mathrm{~N}\). What is the magnitude of the acceleration of the rock?

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