heipCuit w asArialBIUA12三主主手Independent Practice1.A 5000-kg truck traveling with a speed of 10 m/s skids to a stop. The truckexperiences a 20000N force of friction. Determine the stopping distance of the truckComplete Solution stepsWork = Force x distance, KE = % mv?Change in Kinetic Energy = 2(Dm/s)? - V2(m/s)=J.Work Energy Theorem Work done = change in Kinetic Energy

Name: heipCuit w asArialBIUA12三主主手Independent Practice1.A 5000-kg truck tra
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Description: heipCuit w asArialBIUA12三主主手Independent Practice1.A 5000-kg truck traveling with a speed of 10 m/s skids to a stop. The truckexperiences a 20000N force of friction. Determine the stopping distance of the truckComplete Solution stepsWork = Force x di

Answered: 1.A 5000-kg truck traveling with a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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heip
Cuit w as
Arial
BIUA
12
三
主
主
手
Independent Practice
1.A 5000-kg truck traveling with a speed of 10 m/s skids to a stop. The truck
experiences a 20000N force of friction. Determine the stopping distance of the truck
Complete Solution steps
Work = Force x distance, KE = % mv?
Change in Kinetic Energy = 2(
Dm/s)? - V2(
m/s)=
J.
Work Energy Theorem Work done = change in Kinetic Energy

Expert Solution

Step 1

Mass of the truck = 5000 Kg

Its initial velocity = 10 m/s

Its Kinetic energy before skidding= $\frac{1}{2} m v^{2} = \frac{1}{2} \times 5000 \times 10^{2} = 250000 J o u l e s$

Its kinetic energy when it stops after skidding=0 Joules

$∴$ Change in Kinetic Energy =(Final Kinetic Energy-Initial Kinetic Energy) (0-250000)Joule =-250000 Joule

Work = Force X distance= 250000 Joule

$∴ D i s \tan c e = 250000 / 20000 = 12.5 m$

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