/*! 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 26 Under the influence of its drive... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 kg. Under the influence of a drive force of 205 N, it is moving at a constant velocity whose magnitude is 5.50 m/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.

Short Answer

Expert verified
The snowmobile coasts to a halt over a distance of 10.01 meters in approximately 3.64 seconds.

Step by step solution

01

Identify Forces Acting on Snowmobile

The snowmobile moves at a constant velocity under a drive force of 205 N. This implies that the force of friction is equal in magnitude but opposite in direction to the drive force when moving at constant velocity. Hence, the frictional force, when the drive stops, is 205 N.
02

Calculate the Deceleration Due to Friction

Since the drive force stops, only the friction force affects the motion. Using Newton's second law, the deceleration \( a \) is found by \( a = \frac{f}{m} = \frac{205 \text{ N}}{136 \text{ kg}} \approx 1.51 \text{ m/s}^2 \).
03

Use Kinematic Equation to Find Distance

Using the equation \( v^2 = u^2 + 2as \), with final velocity \( v = 0 \), initial velocity \( u = 5.50 \text{ m/s} \), and \( a = -1.51 \text{ m/s}^2 \), solve for distance \( s \). \[ 0 = (5.50)^2 + 2(-1.51)s \] Simplifying, \( s = \frac{(5.50)^2}{2 \times 1.51} \approx 10.01 \text{ meters} \).
04

Calculate Time Using Another Kinematic Equation

Use the kinematic equation \( v = u + at \) to find time \( t \). Again, with \( v = 0 \), \( u = 5.50 \text{ m/s} \), and \( a = -1.51 \text{ m/s}^2 \), solve for \( t \): \[ 0 = 5.50 + (-1.51)t \] Simplification yields \( t = \frac{5.50}{1.51} \approx 3.64 \text{ seconds} \).

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 91Ó°ÊÓ!

Key Concepts

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

kinematic equations
Kinematic equations play a crucial role in solving problems related to motion under uniform acceleration. They are often used in physics to predict the future position or velocity of an object moving with constant acceleration. These equations make it possible to calculate various elements of motion without having to directly measure them.
For the snowmobile, we applied two kinematic equations:
  • To find the distance the snowmobile traveled until it stopped, we used the equation \( v^2 = u^2 + 2as \). Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the deceleration, and \( s \) is the distance. The final velocity is zero because the snowmobile stops.

  • To determine how long it took for the snowmobile to stop, we used the equation \( v = u + at \), which relates the final and initial velocities, acceleration, and time.
Understanding these equations allows one to predict how long it takes for an object to stop under a specific force, and how far it will travel in this timeframe.
frictional force
Frictional force significantly influences motion, especially when a driving force is removed. It originates from the interaction between surfaces in contact and tends to oppose motion, acting opposite to the direction of movement.
In the snowmobile example, when the drive force is cut off, the frictional force becomes the only force acting against its motion. Initially, the frictional force exactly balances the drive force (205 N) when moving at constant velocity. Therefore, when the drive force is halted, the frictional force remains the same and acts alone to stop the snowmobile.
By applying Newton's second law \( F = ma \), where \( F \) is the frictional force, \( m \) is the mass, and \( a \) is acceleration (or deceleration in this case), we can determine the deceleration caused by this frictional force. Understanding the role of friction helps explain why objects do not continue moving indefinitely when no longer propelled by a force.
deceleration
Deceleration is the rate at which an object slows down, which is essentially negative acceleration. It's vital to distinguish between acceleration and deceleration, as they affect motion in opposite ways.
In the context of the snowmobile, deceleration occurs once the drive force is removed, and the opposing frictional force takes over. The deceleration can be calculated by dividing the frictional force by the mass of the snowmobile, resulting in \( a = \frac{205 \text{ N}}{136 \text{ kg}} \approx 1.51 \text{ m/s}^2 \).
Knowing the deceleration helps in predicting how quickly an object will stop. By applying kinematic equations alongside this deceleration value, we can find out not only the distance the snowmobile will coast until it halts but also the time it will take.
Deceleration is a key factor in understanding how forces interact with objects to change their state of motion.

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 2.00-kg rock is released from rest at a height of 20.0 m. Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: 20.0, 10.0, and 0 m.

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is 67 N. What is the magnitude of the pulling force exerted by his wife?

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of 0.75 m. With the string hanging vertically, the object is given an initial velocity of 2.0 m/s parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle \(\theta\) with its initial vertical orientation and then swings back downward. Find the angle \(\theta\) .

The surfer in the photo is catching a wave. Suppose she starts at the top of the wave with a speed of 1.4 m/s and moves down the wave until her speed increases to 9.5 m/s. The drop in her vertical height is 2.7 m. If her mass is 59 kg, how much work is done by the (non conservative) force of the wave?

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 \(4.5 \times 10^{4} kg\) , and the force causes its speed to change from 7100 to 5500 \(m/ s\) . ( a What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} m,\) determine the magnitude of the force.

See all solutions

Recommended explanations on Physics Textbooks

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.