/*! 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 14 A major force opposing the motio... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 14

A major force opposing the motion of a vehicle is the rolling resistance of the tires, \(F_{r}\), given by $$ F_{\mathrm{r}}=f^{\circ} W $$ where \(f\) is a constant called the rolling resistance coefficient and \(W\) is the vehicle weight. Determine the power, in \(\mathrm{kW}\), required to overcome rolling resistance for a truck weighing \(322.5 \mathrm{kN}\) that is moving at \(110 \mathrm{~km} / \mathrm{h}\). Let \(f=0.0069\).

Short Answer

Expert verified
The power required to overcome rolling resistance is 68.009 kW.

Step by step solution

01

Write the Given Values

Identify the values given in the problem:o \(W = 322.5 \text{ kN}\)o \(v = 110 \text{ km/h}\)o \(f = 0.0069\)
02

Calculate the Rolling Resistance Force

Using the formula for rolling resistance force:o \(F_{\text{r}} = f W\)Substitute the given values:o \(F_{\text{r}} = 0.0069 \times 322.5 \text{ kN}\)Calculate the result:o \(F_{\text{r}} = 2.22525 \text{ kN}\)
03

Convert the Speed Unit

Convert the speed from km/h to m/s:o \(v = 110 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}\)Calculate the result:o \(v = 30.56 \text{ m/s}\)
04

Calculate the Power in Watts

Use the formula for power to overcome rolling resistance:o \(P = F_{\text{r}} \times v\)Substitute the calculated force and speed:o \(P = 2.22525 \text{ kN} \times 30.56 \text{ m/s}\)Since 1 kN = 1000 N:o \(P = 2225.25 \text{ N} \times 30.56 \text{ m/s}\)Calculate the result:o \(P = 68009.04 \text{ W}\)
05

Convert Power to Kilowatts

Convert the power from watts to kilowatts:o \(P_{\text{kW}} = \frac{P}{1000}\)Substitute the value of power in watts:o \(P_{\text{kW}} = \frac{68009.04 \text{ W}}{1000}\)Calculate the result:o \(P_{\text{kW}} = 68.009 \text{ kW}\)

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Key Concepts

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

Rolling Resistance Coefficient
Rolling resistance is a key concept in vehicle dynamics. The rolling resistance coefficient (\f\text{\textsuperscript{\text{o}}}\)) represents the frictional force opposing the motion of a vehicle due to the deformation of tires and the surface they travel on.
The formula for calculating the rolling resistance force is:

\( F_{\text{r}} = f^o W \times W \times W \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} \times \frac{1 \text{ MN}}{1000 \text{ kN}} \times \times \times P \times F_{\text{r}} \times v, \times \times P_{\text{kW}} \times P_{\text{kW}} \times 1000 ,$$ \times \frac{1000 \text{ W in}}{1000}$ #00225 is a key concept in vehicle dynamics.

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Most popular questions from this chapter

A storage battery develops a power output of $$ \dot{W}=1.2 \exp (-t / 60) $$ where \(\dot{W}\) is power, in \(\mathrm{kW}\), and \(t\) is time, in s. Ignoring heat transfer (a) plot the power output, in \(\mathrm{kW}\), and the change in energy of the battery, in \(\mathrm{kJ}\), each as a function of time. (b) What are the limiting values for the power output and the change in energy of the battery as \(t \rightarrow \infty\) ? Discuss.

An object whose mass is \(0.5 \mathrm{~kg}\) has a velocity of \(30 \mathrm{~m} / \mathrm{s}\). Determine (a) the final velocity, in \(\mathrm{m} / \mathrm{s}\), if the kinetic energy of the object decreases by \(130 \mathrm{~J}\). (b) the change in elevation, in \(\mathrm{ft}\), associated with a \(130 \mathrm{~J}\) change in potential energy. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

The driveshaft of a building's air-handling fan is turned at 300 RPM by a belt running on a \(0.3\)-m-diameter pulley. The net force applied by the belt on the pulley is \(2000 \mathrm{~N}\). Determine the torque applied by the belt on the pulley, in \(\mathrm{N} \cdot \mathrm{m}\), and the power transmitted, in \(\mathrm{kW}\).

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A gas undergoes a thermodynamic cycle consisting of three processes: Process 1-2: constant volume, \(V=0.028 \mathrm{~m}^{3}, U_{2}-U_{1}=\) \(26.4 \mathrm{~kJ}\) Process 2-3: expansion with \(p V=\) constant, \(U_{3}=U_{2}\) Process 3-1: constant pressure, \(p=1.4 \mathrm{bar}, W_{31}=-10.5 \mathrm{~kJ}\) There are no significant changes in kinetic or potential energy. (a) Sketch the cycle on a \(p-V\) diagram. (b) Calculate the net work for the cycle, in \(\mathrm{kJ}\). (c) Calculate the heat transfer for process 2-3, in kJ. (d) Calculate the heat transfer for process 3-1, in \(\mathrm{kJ}\). Is this a power cycle or a refrigeration cycle?
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