Lesson 1.8: A Closer Look at Braking
Completion requirements
Lesson 1.8: A Closer Look at Braking
Having the ability to bring a vehicle to a stop within a certain distance can prevent a motor vehicle collision. Stopping distance is the sum of the reaction distance and the braking distance. A closer look indicates that braking distance depends on the magnitude of the horizontal force between the tires and the road. During normal braking, the magnitude of this horizontal force increases. However, slippery road conditions can prevent the buildup of this force and can make braking distances dangerously long.This Lessons explores the force of friction and braking. Newtons Second Law of Motion relates acceleration to force and mass
This lesson and assignment should not require much time to complete.
- Read pages 221 and 222 and the first paragraph on page 223. Answer Practice Problem 44 and check your answers with those in the “Practice Answers” in the online course.
Friction is a force that opposes motion by doing work that changes kinetic energy to thermal energy.
Friction is often a problem because useful energy is lost. Cars are only about 35% efficient because of energy lost to air resistance (friction) and friction inside the moving parts of the vehicle. (about 30% of the energy is lost as chemical energy and heat through the exhaust system as well.
Friction is also needed. Friction between the tires and the road allow turning and stopping. Friction between the brake pads and the tire does the work that allows a car to stop.
Net force is the sum of several forces. Net force is often considered in friction problems because friction opposes the work done by another force, so the net force determines the acceleration.
Friction is often a problem because useful energy is lost. Cars are only about 35% efficient because of energy lost to air resistance (friction) and friction inside the moving parts of the vehicle. (about 30% of the energy is lost as chemical energy and heat through the exhaust system as well.
Friction is also needed. Friction between the tires and the road allow turning and stopping. Friction between the brake pads and the tire does the work that allows a car to stop.
Net force is the sum of several forces. Net force is often considered in friction problems because friction opposes the work done by another force, so the net force determines the acceleration.
Understanding Newtons Second Law
Newtons second Law often is written F = ma, as it does in your Data Booklet. But it is best understood as , because the motion of an object (acceleration) depends on two variables; orce and mass.- Force is most easily understood. Apply a large force to your lawnmower, and it will accelerate more quickly than if you apply a small force. The illustration on the right shows mass stays the same, but a small force produces a small acceleration, while a large force produces a large acceleration
Consider the effect of mass. (force will remain the same)
- A puck sitting at rest on an ice surface is easily accelerated with the force from a hockey stick. The illustration shows the same force in both situations. A force on a small mass produces a large acceleration.
- When the Zamboni (the large ice surfacing tractor) is hit with the same amount of force, the acceleration will be very small. the illustration shows how a large mass results in a small acceleration.
In the next activity, you will view "Newtons Second Law" to study factors affecting the change in the volocity of an object. If possible, work with someone who can help you record the data. Stop and start the video as required.
Note: In this activity, the direction of the force is indicated by a sign convention where the positive direction is the direction of the vehicle’s initial velocity. The negative direction is opposite the direction of the initial velocity.
- Read the activity on pages 223 and 224 of the textbook. Follow the directions, and answer the questions. Check your answers with those in the “Suggested Answers” . This investigation is worth spending about 20 minutes on.
- Read page 225 of the textbook. Work through Example Problem 1.21 and do Practice Problem 45. Check your answers with those in the “Practice Answers” .
Net Force
An engine pushes a 1800 kg truck with 3000 N of force. If there is a friction force of 200 N opposing motion, what will be the acceleration of the truck?
From you Data Booklet, Newton's Second Law is: F = ma.
You need to know that the force acceleration the truck is the net force.
- Read “1.8 Summary” on page 226 of the textbook. Doing the “1.8 Questions” on page 227 will be useful. Make sure you do 3bc and 4. Check your answers with those in the “Practice Answers” in the online course.