/*! 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 25 A sled is being pulled across a ... [FREE SOLUTION] | 91Ó°ÊÓ

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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's 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's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

Short Answer

Expert verified
The sled's kinetic energy would increase by approximately 17.84%.

Step by step solution

01

Understand the Initial Scenario

Initially, the force is applied entirely along the \(+x\) axis, causing the sled's kinetic energy to increase by \(38\%\). This means there is no angular component affecting the force.
02

Determine the Work Done in the Initial Scenario

The work done, \( W \), in this case, can be represented by the formula \( W = F \times d \), where \( F \) is the force and \( d \) is the displacement along the \(+x\) axis. This work results in a \(38\%\) increase in kinetic energy \( (\Delta KE) \).
03

Analyze the New Scenario with Angle

When the force is at an angle of \(62^{\circ}\) above the \(+x\) axis, only a component of this force contributes to the work done along the \(x\) axis. This component is \( F_{x} = F \cos(62^{\circ}) \).
04

Calculate the Effective Work Done

The effective work done, which affects the kinetic energy change, is now given by \( W' = F \cos(62^{\circ}) \times d \). This represents the new work contributing to the kinetic energy change.
05

Relate the Work Done to Kinetic Energy Increase

The new change in kinetic energy \((\Delta KE')\) is proportional to the work done \((W')\). Thus, \( \Delta KE' = \Delta KE \times \cos(62^{\circ}) \).
06

Compute the Percentage Increase in Kinetic Energy

Given the original \(38\%\) increase, the new increase is \(38\% \times \cos(62^{\circ}) = 38\% \times 0.4695 \approx 17.84\%\). Therefore, the kinetic energy would increase by approximately \(17.84\%\).

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Key Concepts

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

Work-Energy Principle
The Work-Energy Principle connects the concepts of work and kinetic energy. When work is done on an object, it results in a change in its kinetic energy. This principle is expressed mathematically as \( W = \Delta KE \), where \( W \) is the work done and \( \Delta KE \) is the change in kinetic energy.

In our scenario with the sled, the force applied along the same direction as the sled's motion amplifies its kinetic energy by 38%. This means all the work done by the force effectively converts into an increase in kinetic energy.

When work is performed, energy is transferred or transformed. If the pulling force is entirely in the direction of the sled's movement, the maximum amount of work is converted into kinetic energy.
Angle of Force Application
The angle at which a force is applied plays a critical role in determining how much of the force contributes to the desired motion. In the original exercise, when the force was applied directly along the direction of sled's motion, it led to a 38% increase in kinetic energy.

However, when the force is at an angle, not all of it contributes to moving the sled forward. For instance, a force at \( 62^{\circ} \) above the horizontal means that only the horizontal component of the force is effective in doing work along the sled's path. This component is calculated using \( F_{x} = F \cos(62^{\circ}) \).

The effective work done, which changes the sled's kinetic energy, is now reduced, demonstrating how angle affects the work-energy relationship.
Force Components
When a force is applied at an angle, it can be broken into components to understand how it affects motion in different directions. The force along the x-axis (directly aiding the sled's forward motion) is \( F_{x} = F \cos(62^{\circ}) \), while the vertical component is \( F_{y} = F \sin(62^{\circ}) \).

In terms of increasing kinetic energy, only the horizontal component \( F_{x} \) is relevant.
  • The horizontal component directly influences the sled’s velocity as it is parallel to the sled's trajectory.
  • The vertical component does not affect the horizontal kinetic energy but may influence other forces like normal force.


Understanding these components helps clarify why the sled's kinetic energy increase is less significant when the force is not directly applied in the direction of movement.

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Most popular questions from this chapter

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of \(4.5 \times 10^{4} \mathrm{kg},\) and the force causes its speed to change from 7100 to \(5500 \mathrm{m} / \mathrm{s}\). (a) What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} \mathrm{m},\) determine the magnitude of the force.

The concepts in this problem are similar to those in MultipleConcept Example \(4,\) except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 79 -kg person straight up by means of a cable. The person has an upward acceleration of \(0.70 \mathrm{m} / \mathrm{s}^{2}\) and is lifted from rest through a distance of \(11 \mathrm{m}\). (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work-energy theorem and find the final speed of the person.

Multiple-Concept Example 5 reviews many of the concepts that play roles in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of \(25.0^{\circ}\) with the horizontal. The coefficient of kinetic friction between her skis and the snow is \(0.200 .\) She coasts down a distance of \(10.4 \mathrm{m}\) before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is \(3.50 \mathrm{m}\) below the edge. How fast is she going just before she lands?

A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of \(14.6^{\circ}\) with the horizontal. What is the change in the skier's gravitational potential energy?

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of \(136 \mathrm{kg}\). Under the influence of a drive force of \(205 \mathrm{N},\) it is moving at a constant velocity whose magnitude is \(5.50 \mathrm{m} / \mathrm{s}\) The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

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