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Automotive Power I. A truck engine transmits 28.0 kW (37.5

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 100P Chapter 6

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 100P

Automotive Power I. A truck engine transmits 28.0 kW (37.5 hp) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 km/h (37.3 mi/h) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 km/h? At 120.0 km/h? Give your answers in kilo watts and in horsepower.

Step-by-Step Solution:

Solution 100P Step 1: Data given Power 280 KW = 28000 W Velocity v = 60.0 km /hr = 16.67 m/s We need to find the resisting force acting on the truck We have relation P = F × v Using this we shall obtain the force acting on the truck F = P/v Substituting values we get F = 28000 W/16.67 m/s F = 1679.6 N This can be approximated to 1680 N Hence the total resistance force acting on the truck is 1680N Step 2 : It is given 65% of the resistance force is due to rolling friction That is FR= 1092 N is rolling force And 35% is due to air resistance That is FA= 588 N is air friction force v2= 30 km /h = 8.33 m/s v3= 120.0 km /h = 33.33 m/s We need to find power at velocity v 2 We shall find coefficient of air friction force F = kv 2 A k = F Av 2 Substituting values we get k = 588 N/(16.67 m/s) 2 k = 2.11 For velocity v2friction due to air is 2 F A k × v 2 Substituting values we get 2 F A 2.12 × (8.33 m/s) F A 147 N Since the rolling friction is constant The total resistive will be F = F + F = 1092 N + 147N R A F = 1239 N Using this we have power as P = F × v Substituting values we get P = 1239 N × 8.33 m/s = 13320.87 W This can be written as P = 10.3 KW In terms of horsepower this is written as 1.34 Hp P = 10.3 KW × 1KW = 13.81 Hp Thus for velocity v2we have power as 10.3 KW = 13.81 Hp

Step 3 of 3

Chapter 6, Problem 100P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

Since the solution to 100P from 6 chapter was answered, more than 1776 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 100P from chapter: 6 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “Automotive Power I. A truck engine transmits 28.0 kW (37.5 hp) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 km/h (37.3 mi/h) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 km/h? At 120.0 km/h? Give your answers in kilo watts and in horsepower.” is broken down into a number of easy to follow steps, and 110 words. This full solution covers the following key subjects: Force, TRUCK, resistance, rolling, resisting. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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Automotive Power I. A truck engine transmits 28.0 kW (37.5