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Chapter 5: Problem 61

A car does the work \(W_{\text {car }}=7.0 \cdot 10^{4} \mathrm{~J}\) in traveling a distance \(x=2.8 \mathrm{~km}\) 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

We are given the distance traveled by the car in kilometers (\(x=2.8\,\text{km}\)). To solve the problem, we need the distance in meters. To convert kilometers to meters, multiply the distance by 1000: \(x = 2.8\,\text{km} \times 1000 = 2800\,\text{m}\)
02

Use the work-energy theorem

Since the car is moving at a constant speed, there is no change in kinetic energy, and we can use the equation for work done, \(W=F \cdot d \cdot \cos \theta\). Since we assume the angle \(\theta\) to be \(0^{\circ}\), we have \(\cos \theta = 1\). Thus, the equation simplifies to: \(W = F \cdot d\)
03

Solve for the average force

Now, we have all the information needed to solve for the average force \(F\). We know the work done by the car is \(W_{\text{car}} = 7.0 \cdot 10^4 J\), and the distance traveled is \(x=2800 m\). Using the equation \(W=F \cdot d\), we can solve for the force: \(F = \frac{W}{d} = \frac{7.0 \cdot 10^4\,\text{J}}{2800\,\text{m}}\)
04

Calculate the average force

Now, perform the calculation to find the average force acting on the car: \(F = \frac{7.0 \cdot 10^4\,\text{J}}{2800\,\text{m}} = 25\,\text{N}\) Thus, the average force acting on the car during this process is \(F = 25\,\text{N}\).

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Most popular questions from this chapter

A car of mass \(1214.5 \mathrm{~kg}\) is moving at a speed of \(62.5 \mathrm{mph}\) when it misses a curve in the road and hits a bridge piling. If the car comes to rest in \(0.236 \mathrm{~s}\), how much average power (in watts) is expended in this interval?

Which of the following is a correct unit of power? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) J e) \(W\) b) \(N\) d) \(\mathrm{m} / \mathrm{s}^{2}\)

A spring with spring constant \(k\) is initially compressed a distance \(x_{0}\) from its equilibrium length. After returning to its equilibrium position, the spring is then stretched a distance \(x_{0}\) from that position. What is the ratio of the work that needs to be done on the spring in the stretching to the work done in the compressing?

Does the Earth do any work on the Moon as the Moon moves in its orbit?

A certain tractor is capable of pulling with a steady force of \(14 \mathrm{kN}\) while moving at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). How much power in kilowatts and in horsepower is the tractor delivering under these conditions?
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