Calculate Force & Coefficient of Friction for 2000kg Car & 1.2 x 10^8 J/Gallon

[pippintook]

A gallon of gasoline has a chemical PE of 1.2 x 10^8 J. If a 2000 kg car gets 40 km/gallon, what is the average force on the car? Assume constant speed and remember the force on the car is applied over the total distance traveled. What is the coefficient of friction?

I tried PE = mgx, using m = 2000 and solving for x (which was 6122.45). I'm not really sure where to go from there though.

 

[LowlyPion]

Not quite.

The PE of the gasoline is not gravitational PE. Completely different.

This PE gets converted into work over the distance of 40 km.

What you do is convert the energy they gave you into the work to move the car. Work is F*d.

At constant speed then the force retarding motion is consumed by the combustion of the gas.

 

[thomate1]

To calculate the average force on the car, we can use the formula F = ma, where F is the force, m is the mass of the car, and a is the acceleration. Since we are assuming constant speed, the acceleration is 0. Therefore, the average force on the car is 0 N.

To calculate the coefficient of friction, we need to use the formula Ff = μN, where Ff is the force of friction, μ is the coefficient of friction, and N is the normal force. In this case, the normal force is equal to the weight of the car, which is mg. So, we can rewrite the formula as Ff = μmg.

Since we know the mass of the car (2000 kg) and the force of gravity (9.8 m/s^2), we can calculate the normal force to be 19600 N.

Now, we need to find the force of friction. We can do this by using the equation Ff = Wd, where W is the work done and d is the distance traveled. In this case, the work done is equal to the chemical potential energy of the gasoline, which is 1.2 x 10^8 J. The distance traveled is equal to the number of gallons of gasoline used, which is equal to 40 km/gallon. Therefore, the distance traveled is 40 km x 1000 m/km = 40000 m.

Plugging in these values, we get Ff = 1.2 x 10^8 J x 40000 m = 4.8 x 10^12 J.

Now, we can plug this value into our equation for force of friction and solve for the coefficient of friction:

4.8 x 10^12 J = μ x 19600 N

μ = (4.8 x 10^12 J) / (19600 N) = 2.45 x 10^8

Therefore, the coefficient of friction for this car is 2.45 x 10^8. This is a very high value, which suggests that there is a lot of friction between the tires and the road, possibly due to worn out tires or rough road conditions.