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A car does the work \(W_{c a r}=7.00 \cdot 10^{4} \mathrm{~J}\) in traveling a distance \(x=2.80 \mathrm{~km}\) at constant speed. Calculate the average force \(F\) (from all sources) acting on the car in this process.

Short Answer

Expert verified
Answer: The average force acting on the car is 25 N.

Step by step solution

01

Convert distance to meters

First, we need to convert the distance given in kilometers to meters. To do this, we will use the following conversion factor: 1 km = 1000 m. So, the distance in meters will be: $$ x = 2.80 \thinspace km \times 1000 = 2800 \thinspace m $$
02

Write down the formula for work done

We will use the work-energy principle to find the average force acting on the car. The formula for work done is given by: $$ W_{car} = F \times x $$
03

Rearrange the formula to solve for force

To calculate the average force, we will rearrange the formula in Step 2 to solve for \(F\): $$ F = \frac{W_{car}}{x} $$
04

Plug in the values and calculate the average force

Now, we can substitute the values of \(W_{car}\) and \(x\) into the equation above to find the average force \(F\) acting on the car: $$ F = \frac{7.00 \cdot 10^4 \thinspace J}{2800 \thinspace m} = 25 \thinspace N $$ So, the average force acting on the car during this process is 25 Newtons.

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Key Concepts

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

Physics Work Calculation
Understanding physics work calculation is an essential part of understanding how objects interact and apply forces to each other, leading to movement or change in energy states. Work is defined as the product of the force applied to an object and the displacement of that object in the direction of the force. This relationship is expressed mathematically as:

\[\begin{equation}W = F \times d\right), \text{where }W\text{ is work, }F\text{ is the constant force applied, and }d\text{ is the displacement.}\right)\right), \end{equation}\]In the case of the car problem, it’s vital for students to understand that only the component of force acting in the direction of the car’s displacement does work. If the car is moving at a constant speed, this means the net work done by all forces (like gravity, friction, etc.) is zero, but the car still does work against these forces to maintain its speed.
Average Force
Average force is a conceptual tool that allows us to simplify complex scenarios where the force may vary throughout the process. It is the constant force that would be required to produce the same effect (such as displacement) on an object as a series of actual varying forces over a given time period or distance. To calculate the average force, you use the formula derived from the work-energy principle.

In the exercise, once we've established the work done by the car, we rearrange the work formula to solve for the average force (F\right): \[\begin{equation}F = \frac{W}{d}\right), \end{equation}\]which yields the average force over the given distance. It's a powerful way to simplify the problem while maintaining the essence of the physics involved.
Conversion of Units
Physics problems often involve working with a variety of units, so knowing how to properly convert these units is crucial. Unit conversion ensures that all terms in an equation are consistent and can properly interact. For instance, in our solution, we convert the distance from kilometers to meters since the standard unit for work (joules) involves meters, not kilometers. This ensures consistency with the standard unit of force in newtons, which is derived from joules and meters (1 joule is the work done by a force of 1 newton moving an object 1 meter).

Proper conversion involves multiplying by a conversion factor, which is a way of expressing the equivalent amount of one unit as another. In our exercise, the conversion factor is 1 kilometer equals 1000 meters. Failure to convert units correctly can lead to incorrect results and confusion, especially in more complex physics calculations.

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Most popular questions from this chapter

An \(800-\mathrm{N}\) box is pushed up an inclined plane that is \(4.0 \mathrm{~m}\) long. It requires \(3200 \mathrm{~J}\) of work to get the box to the top of the plane, which is \(2.0 \mathrm{~m}\) above the base. What is the magnitude of the average friction force on the box? (Assume the box starts at rest and ends at rest.) a) zero b) not zero but less than \(400 \mathrm{~N}\) c) greater than \(400 \mathrm{~N}\) d) \(400 \mathrm{~N}\) e) \(800 \mathrm{~N}\)

Santa's reindeer pull his sleigh through the snow at a speed of \(2.561 \mathrm{~m} / \mathrm{s}\) The mass of the sleigh, including Santa and the presents, is \(540.3 \mathrm{~kg} .\) Assuming that the reindeer can provide a total power of \(2.666 \mathrm{hp}\), what is the coefficient of friction between the runners of the sleigh and the snow?

A force has the dependence \(F_{x}(x)=-k x^{4}\) on the displacement \(x\), where the constant \(k=20.3 \mathrm{~N} / \mathrm{m}^{4}\). How much work does it take to change the displacement, working against the force, from \(0.730 \mathrm{~m}\) to \(1.35 \mathrm{~m} ?\)

The mass of a physics textbook is \(3.40 \mathrm{~kg}\). You pick the book up off a table and lift it \(0.470 \mathrm{~m}\) at a constant speed of \(0.270 \mathrm{~m} / \mathrm{s}\). a) What is the work done by gravity on the book? b) What is the power you supplied to accomplish this task?

A particle of mass \(m\) is subjected to a force acting in the \(x\) -direction. \(F_{x}=(3.00+0.500 x) \mathrm{N} .\) Find the work done by the force as the particle moves from \(x=0.00\) to \(x=4.00 \mathrm{~m}\)

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