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An engine expends \(40.0\) hp in propelling a car along a level track at a constant speed of \(15.0 \mathrm{~m} / \mathrm{s}\). How large is the total retarding force acting on the car? Remember that \(1 \mathrm{hp}=745.7 \mathrm{~W}\).

Short Answer

Expert verified
The total retarding force on the car is 1988.53 N.

Step by step solution

01

Convert horsepower to watts

The power given is in horsepower (hp), and we need to convert it to watts (W) to use it in our calculations. Since 1 hp is equivalent to 745.7 W, we have \(40.0 \text{ hp} = 40.0 \times 745.7 \text{ W}\). Calculating this gives \(29828 \text{ W}\).
02

Understand the relationship between power, force, and speed

Power (P) is related to force (F) and speed (v) by the equation \(P = F \times v\). In this equation, P is in watts, v is in meters per second, and F is the force in newtons that we need to find.
03

Rearrange the equation to solve for force

Rearrange the equation from the previous step to solve for force: \(F = \frac{P}{v}\).
04

Substitute the known values into the equation

Now, substitute the known values: \(P = 29828 \text{ W}\) and \(v = 15.0 \text{ m/s}\) into the equation. This yields \(F = \frac{29828}{15.0}\).
05

Calculate the retarding force

Perform the division to find the force: \(F = 1988.53 \text{ N}\). This is the total retarding force acting on the car.

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Key Concepts

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

Power conversion
When dealing with engines or machines, an important concept is power conversion. Power is the rate at which energy is used or transferred. In this exercise, an engine's power is given in horsepower (hp), a common unit used for engines. However, in physics, we often use watts (W) as the unit of power.
To solve problems in physics, converting horsepower to watts is necessary because it aligns with standard SI units. The conversion factor between horsepower and watts is critical: \(1 \text{ hp} = 745.7 \text{ W}\). Therefore, for an engine exerting \(40.0 \text{ hp}\), we calculate the equivalent power in watts by multiplying: \(40.0 \times 745.7 = 29828 \text{ W}\).
This conversion allows us to work within a consistent unit framework, making it easier to apply formulas and understand the quantities involved. Consistent unit systems simplify equations and help avoid mistakes.
Force calculation
Force calculation is a fundamental step in understanding how much interaction is necessary to produce a particular effect, such as moving a car at a steady speed. In this situation, the engine exerts power to overcome a retarding force keeping the car moving on a flat track.
The relationship between power, force, and speed is given by the formula \(P = F \times v\). This shows the direct link between how fast power is used and the amount of force that results. To determine the force, we can rearrange this formula to \(F = \frac{P}{v}\).
By inputting the known power in watts and speed in meters per second, \(P = 29828 \text{ W}\) and \(v = 15.0 \text{ m/s}\), we find: \(F = \frac{29828}{15.0} = 1988.53 \text{ N}\). This force is what we call the retarding force, counteracting the car's motion. Calculating force in this way helps us understand the mechanical demands of maintaining motion.
Kinematics
Kinematics involves studying motion without considering the forces that cause this motion, though the forces can still be part of the overall problem, as they often are in physics problems like this one.
In the given exercise, although we primarily focus on force and power, understanding the car's constant speed is a key kinematic consideration. Constant speed implies that the acceleration is zero. Therefore, the retarding force (including friction or air resistance) is exactly balanced by the engine's force.
This balance is a classic problem in kinematics showing that motion is maintained by this force equilibrium. When speed is constant, forces are balanced: the driving force exerted by the engine perfectly counteracts the retarding forces. Understanding kinematics in this context helps to logically deduce how forces interact to maintain this uniform motion.

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Most popular questions from this chapter

A 50000 -kg freight car is pulled \(800 \mathrm{~m}\) up along a \(1.20\) percent grade at constant speed. (a) Find the work done against gravity by the draw- bar pull. (b) If the friction force retarding the motion is \(1500 \mathrm{~N}\), find the total work done.

A \(2.0\) -kg mass falls \(400 \mathrm{~cm}\). (a) How much work was done on it by the gravitational force? ( \(b\) ) How much \(\mathrm{PE}_{\mathrm{G}}\) did it lose? (c) Given that work is the transfer of energy, where does that energy end up? (a) Gravity pulls with a force \(m g\) on the object, and the displacement is \(4 \mathrm{~m}\) in the direction of the force. The positive work done by gravity is therefore $$ (m g)(4.00 \mathrm{~m})=(2.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(4.00 \mathrm{~m})=78 \mathrm{~J} $$ (b) The change in \(\mathrm{PE}_{\mathrm{G}}\) of the object is \(m g h_{f}-m g h_{i}\), where \(h_{i}\) and \(h_{f}\) are the initial and final heights of the object above the reference level. We then have $$ \begin{array}{l} \text { Change in } \mathrm{PE}_{\mathrm{G}}=m g h_{f}-m g h_{i}=m g\left(h_{f}-h_{i}\right)-(2.0 \mathrm{~kg} \times 9.81 \\ \mathrm{N})(-4.0 \mathrm{~m})=-78 \mathrm{~J} \end{array} $$ The loss in \(\mathrm{PE}_{\mathrm{G}}\) is \(78 \mathrm{~J}\). (c) Gravity provides the force that accelerates the \(2.0\) -kg mass and increases its kinetic energy by \(78 \mathrm{~J}\).

A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

Water flows from a reservoir at the rate of \(3000 \mathrm{~kg} / \mathrm{min}\), to a turbine \(120 \mathrm{~m}\) below. If the efficiency of the turbine is 80 percent, compute the power output of the turbine. Neglect friction in the pipe and the small KE of the water leaving the turbine. Don't forget that it's only 80 percent efficient.

A moving 300 -g object slides unpushed \(80 \mathrm{~cm}\) in a straight line along a horizontal tabletop. How much work is done in overcoming friction between the object and the table if the coefficient of kinetic friction is \(0.20\) ? First find the friction force. Since the normal force equals the weight of the object, $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(0.300 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.588 \mathrm{~N} $$ The work done overcoming friction is \(F_{\mathrm{f}} S \cos \theta\). Here \(\theta\) is the angle between the force and the displacement. Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ}\). Therefore, Work \(=F_{\mathrm{f}} S \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J}\) The work is negative because the friction force is oppositely directed to the displacement; it slows the object and it decreases the object's kinetic energy, or more to the point, it opposes the motion.

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