/*! 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 68 An iceboat is at rest on a frict... [FREE SOLUTION] | 91Ó°ÊÓ

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An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of \(200 \mathrm{~N},\) toward the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of \(8.0 \mathrm{~m}\) in a direction \(20^{\circ}\) north of east. What is the kinetic energy of the iceboat at the end of that \(8.0 \mathrm{~m} ?\)

Short Answer

Expert verified
Kinetic energy is 1503.52 J.

Step by step solution

01

Identify Key Information

We have an iceboat initially at rest, a force of \(200 \, \text{N}\) applied, and the boat moving \(8.0 \, \text{m}\) in a direction \(20^\circ\) north of east. We need to find the kinetic energy of the iceboat after it travels this distance.
02

Understand Work-Energy Principle

The work-energy principle states that the work done by the force on the object results in a change in the kinetic energy. The formula is given by \( W = \Delta KE \). Our goal is to calculate the work done because this will directly give us the change in kinetic energy since the boat starts at rest.
03

Calculate the Work Done

The work done \( W \) by a force \( F \) when it moves an object through a displacement \( d \) at an angle \( \theta \) is \( W = F \, d \, \cos(\theta) \). Here, \( F = 200 \, \text{N} \), \( d = 8.0 \, \text{m} \), and \( \theta = 20^\circ \). Thus, \[ W = 200 \, \text{N} \times 8.0 \, \text{m} \times \cos(20^\circ) \].
04

Perform the Calculations

First, calculate \( \cos(20^\circ) \), and then substitute all the values into the work formula: \[ \cos(20^\circ) \approx 0.9397 \]. \[ W = 200 \, \text{N} \times 8.0 \, \text{m} \times 0.9397 = 1503.52 \, \text{J} \].
05

Conclude the Kinetic Energy

Since the work done is equal to the change in kinetic energy (starting from rest), the kinetic energy of the iceboat at the end of the \(8.0 \, \text{m}\) is \(1512.94 \, \text{J}\).

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Key Concepts

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

Work-Energy Principle
In physics, the work-energy principle is fundamental to understanding how forces affect the motion of objects. This principle states that the work done by a force on an object is equal to the change in the object's kinetic energy. In simpler terms, when you apply a force to an object over a certain distance, you're doing work, which transforms into the object's motion energy, or kinetic energy.

Here's the basic thought: if an object starts from rest and a force acts upon it, the energy transferred to the object (through work) will change its kinetic energy. Whether the force speeds up, slows down, or stops the object depends on how much work is done and in what direction.
  • If the object starts at rest, then the work done on it directly corresponds to its final kinetic energy.
  • If it wasn't at rest, the change in kinetic energy would be the difference between the final and initial energy states.


For our iceboat, beginning from rest means that all work done by the wind's force transforms into kinetic energy, making calculations neat and straightforward. Overall, this principle helps us bridge the gap between force, motion, and energy transformation.
Angle of Force Application
The angle at which a force is applied significantly affects the amount of work done on an object. When calculating work done by a force, the angle plays a vital role because it determines how much of that force contributes to the actual movement of the object in the desired direction.

The mathematical expression for work, \( W = F \, d \, \cos(\theta) \), highlights this concept. The force \( F \) is applied over a displacement \( d \), and \( \theta \) is the angle between the force and the direction of movement.
  • If the angle \( \theta = 0^\circ \), the entire force contributes to the work done (since \( \cos(0^\circ) = 1 \)).
  • At \( 90^\circ \), no work is done (because \( \cos(90^\circ) = 0 \)), meaning the force is perpendicular to the direction of motion.
  • Any angle in between means only part of the force affects the motion.


In the iceboat scenario, the force from the wind acts at a \( 20^\circ \) angle north of east. This means a significant portion of the force propels the boat forward, but some of the force contributes to lateral movement. Utilizing the cosine factor helps isolate the effective component of the force in the direction of displacement, ensuring accurate work and energy calculations.
Displacement in Physics
Displacement is a vector quantity that refers to the change in position of an object. It’s distinct from distance, as displacement considers both the magnitude and direction of movement. In physics, displacement is crucial as it helps quantify how far and in what direction an object moves from its starting point.

When we talk about displacement, it is essential to remember:
  • It deals with final vs. initial position, not the path taken.
  • The direction component of displacement affects how forces applied influence work done and movement.
  • Even if an object travels a long distance in a curve, the displacement focuses on the straight-line difference between start and finish points.


For the iceboat, the displacement is 8.0 meters in a direction 20 degrees north of east. This displacement tells us exactly how far and in what direction the iceboat travels due to the force. The precise measurement and direction then allow us to plug into the work-energy calculations to understand how energy was transferred due to this movement. Understanding displacement helps clarify how forces produce motion in a linear and directed manner.

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Most popular questions from this chapter

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