/*! 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 79 A water-skier, moving at a speed... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 79

A water-skier, moving at a speed of \(9.30 m/ s,\) is being pulled by a tow rope that makes an angle of \(37.0^{\circ}\) with respect to the velocity of the boat (see the drawing). The tow rope is parallel to the water. The skier is moving in the same direction as the boat. If the tension in the tow rope is \(135 N,\) determine the work that it does in 12.0 \(s\) .

Short Answer

Expert verified
The work done by the tension in the tow rope is approximately 12020.88 Joules.

Step by step solution

01

Understanding the Problem

To determine the work done by the tension in the tow rope, we need to calculate how much of the tension contributes to moving the skier in the direction of motion. This involves finding the component of the tension force that acts in the direction of the skier's motion.
02

Calculate the Horizontal Component of Force

Since the rope makes an angle of \( 37.0^{\circ} \) with the skier's direction, use the cosine of this angle to find the horizontal component of the tension force: \( F_x = 135 \text{ N} \times \cos(37.0^{\circ}) \).
03

Substitute and Calculate

Using a calculator, find \( \cos(37.0^{\circ}) \approx 0.7986 \). Therefore, \( F_x = 135 \times 0.7986 \approx 107.8 \text{ N} \).
04

Find the Distance Traveled

To find the distance the skier travels over 12 seconds at a speed of \( 9.30 \text{ m/s} \), use the formula: \( \text{distance} = \text{speed} \times \text{time} = 9.30 \times 12.0 \).
05

Substitute and Calculate Distance

Calculate the distance: \( 9.30 \times 12.0 = 111.6 \text{ meters} \).
06

Calculate the Work Done

Work done by the tension force is given by \( \text{Work} = F_x \times \text{distance} \). Substitute the values: \( \text{Work} = 107.8 \; \text{N} \times 111.6 \; \text{m} \).
07

Substitute and Final Calculation

Calculate the work: \( \text{Work} = 107.8 \times 111.6 \approx 12020.88 \text{ J} \).

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.

Tension Force
A tension force is the pulling force transmitted through a string, rope, or cable when it is pulled tight by forces acting at each end. In the context of the water-skier problem, the tension force is exerted by the tow rope pulling the skier forward.
To thoroughly understand this:
  • Components of Tension: Tension in a rope can be broken down into two components: horizontal and vertical. However, since the tow rope is parallel to the water, the vertical component doesn't impact the skier's motion.
  • Direction of Tension: Only the horizontal component contributes to moving the skier forward. The rest provides no forward propulsion.
  • Using Angles: In this case, the tension is affected by the angle the rope makes with the boat's velocity. Calculating this component helps us understand how effectively the force is aiding the skier's motion.
Trigonometric Functions in Physics
Trigonometry in physics helps us understand how forces work when they don鈥檛 act along straightforward axes (x, y, or z). Using trigonometric functions like sine, cosine, and tangent, we can estimate real-world components of forces.
For this skier problem:
  • Cosine Function: When the rope makes an angle with the direction of motion, you use the cosine of the angle to find how much of the tension force acts in the desired direction.
  • Understanding Cosine: The cosine of an angle in a triangle gives you the ratio of the length of the adjacent side to the hypotenuse. Here, it converts total tension to horizontal force.
  • Applying the Equation: Thus, applying the formula: \[ F_x = F imes \cos(\theta) \] where \( F \) is the tension force and \( \theta \) is the angle, you accurately determine the effective force involved.
Kinematics in One Dimension
Kinematics deals with the motion of objects without considering the forces that cause the motion. In this exercise, we're interested in one-dimensional motion, which simplifies many calculations.
  • Simple Motion: Since the skier moves in a straight line, we only consider one direction. Here, the crucial factors are speed and time, giving us distance.
  • Displacement and Time: Given that speed is constant, the traveled distance over a period of time is straightforward to calculate: \[ \text{distance} = \text{speed} \times \text{time} \] Using the formula, this step helps us find how far the skier has traveled.
  • Application in Work Calculation: The calculated distance contributes to finding the total work done by the force, illustrating an important link between kinematics and work-energy.
By understanding both the principles of trigonometry and kinematics, you can effectively solve the exercise on tension force acting on the skier.

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 pole-vaulter just clears the bar at 5.80 m and falls back to the ground. The change in the vaulter鈥檚 potential energy during the fall is \(-3.70 \times 10^{3} J .\) What is his weight?

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 \(m/ s\) . After coasting 1.9 \(m\) up the slope, the skier has a speed of 4.4 \(m/s\) . (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?

鈥淩ocket Man鈥� has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and accelerates straight upward. At a height of 16 m, his speed is 5.0 m/s. His mass, including the propulsion unit, has the approximately constant value of 136 kg. Find the work done by the force generated by the propulsion unit.

A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled鈥檚 displacement, which is along the x axis. As a result, the kinetic energy of the sled increases by 38%. By what percentage would the sled鈥檚 kinetic energy have increased if this force had pointed \(62^{\circ}\) above the x axis?

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by seven-time-winner Lance Armstrong (m 75.0 kg) is 6.50 W per kilogram of his body mass. (a) How much work does he do during a 135-km race in which his average speed is 12.0 m/s? (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Thermodynamics

Read Explanation

Engineering Physics

Read Explanation

Particle Model of Matter

Read Explanation

Waves Physics

Read Explanation

Dynamics

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.