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 \(F = 25\,\text{N}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free