Energy and acceleration

[brake4country]

Homework Statement

The chemical potential energy in gasoline is converted to kinetic energy in cars. If a car accelerates from zero to 60 km/h, compared to the energy necessary to increase the velocity of the car from zero to 30 km/h the energy necessary to increase the velocity of the car from 30 to 60 km/h is:

Homework Equations

K = 1/2 mv^2

The Attempt at a Solution

I approached this problem in a conceptual manner. If we are to compare the energy from 0 to 60, it wouldn't matter if we go 0 to 30 or 30 to 60 since acceleration is not factored in. Thus it would be the same. However, the correct answer is 3 times as great. How is this possible?

 

[DrClaude]

I approached this problem in a conceptual manner. If we are to compare the energy from 0 to 60, it wouldn't matter if we go 0 to 30 or 30 to 60 since acceleration is not factored in. Thus it would be the same. However, the correct answer is 3 times as great. How is this possible?

You have to compare the energy required to bring the car from 0 km/h to 30 km/h, versus the energy required to take it from 30 km/h to 60 km/h.

 

[brake4country]

So, that would be K = 1/2 m (30)^2 = 450 m; and then K = 1/2 m (30)^2 = 450 m. This makes no sense!

 

[DrClaude]

So, that would be K = 1/2 m (30)^2 = 450 m; and then K = 1/2 m (30)^2 = 450 m. This makes no sense!

Of course, if you're calculating the same thing twice, you get the same number

I guess that you are putting in the second equation (60 - 30)² for the velocity, but that is not correct. What you need to calculate is the difference in kinetic energy,
$$
K(v = 30\ \mathrm{km/h}) - K(v = 0\ \mathrm{km/h})
$$
compared to $$
K(v = 60\ \mathrm{km/h}) - K(v = 30\ \mathrm{km/h})
$$

To be clear, $\Delta K \neq \frac{1}{2} m (\Delta v)^2$.

 

[HalfLostAndSearching]

I would like to clarify and expand upon the concepts presented in the question. The conversion of chemical potential energy to kinetic energy in a car is a result of the combustion of gasoline, which releases energy in the form of heat. This heat energy is then converted into mechanical energy, which is used to move the car.

In order to accelerate the car, the engine must apply a force to overcome the car's inertia and increase its velocity. This requires a certain amount of energy, which can be calculated using the equation K = 1/2 mv^2, where K is the kinetic energy, m is the mass of the car, and v is the velocity.

Now, let's consider the two scenarios presented in the question: increasing the car's velocity from 0 to 30 km/h and from 30 to 60 km/h. In both cases, the car is starting from rest and must be accelerated to the desired velocity. However, in the second scenario, the car is already moving at 30 km/h, which means it has some kinetic energy from the previous acceleration. Therefore, the energy required to increase its velocity from 30 to 60 km/h is not the same as the energy required to increase it from 0 to 30 km/h.

To calculate the energy required in the second scenario, we need to consider the change in kinetic energy, which is equal to the difference between the final kinetic energy (at 60 km/h) and the initial kinetic energy (at 30 km/h). This can be written as follows:

ΔK = 1/2 m(60^2) - 1/2 m(30^2)

= 1/2 m(3600 - 900)

= 1/2 m(2700)

= 1350 m

On the other hand, the energy required in the first scenario can be calculated as follows:

K = 1/2 mv^2

= 1/2 m(30^2)

= 1/2 m(900)

= 450 m

Therefore, the energy required to increase the car's velocity from 30 to 60 km/h is 3 times greater than the energy required to increase it from 0 to 30 km/h. This is because in the second scenario, the car already has some kinetic energy that needs to be added to its total kinetic energy, while in the first scenario, all of the energy