Can you check my work?

[heikediguore]

I wanted to find out how much food energy it would take to run a car. Thought experiment:

 

Starting from a stop, a 1000 kg car goes 5 km in 5 min. How many 2 oz. Snickers bars does it take to provide this energy?

 

 

5 min. = 300 s

5 km = 5000 m

so velocity = ~17 m/s

 

Energy = E(final) - E(initial)

= (1/2 * mv^2) - 0

= 1/2*(1000 kg)(17 m/s)^2

= 289,000 kg (m/s)^2

= 289,000 j

 

so power requirement is:

289,000 j/s

=289,000 W

 

sustained over 300 sec,

Energy = power * time

=289,000 W * 300 s

=8.67 * 10^7 j

 

1 calorie = 4.184 joules

(8.67 * 10^7 j)/(4.184 j/cal)

=2.07 * 10^7 cal

 

1000 cal = 1 kcal

(2.07 * 10^7 cal)/(1000 cal/kcal)

=2.07 * 10^4 kcal

 

1 Snickers bar = 273 kcal

(see http://www.thecaloriecounter.com/Foods/1900/19155/1/Food.aspx)

 

2.07 * 10^4 kcal/(273 kcal/Snickers)

=75.8 Snickers bars

 

Is that correct?

[timo]

I don't think it's correct.

 

- The amount of energy required to cross 5 km in 5 minutes with a car is not unique. As you probably know, fuel consumption of a car depends on driving style. Notably on the driving speed. Just race close to the destination and then take a break until time ran up. You'll have spent more fuel than if you had been driving slower and -due to having a break- still can need the same time. Did you possibly mean the energy required to accelerate a car to a speed of 5 km per 5 minutes? That would greatly simplify your calculation and make the part where you seem to have made a huge error unnecessary.

 

- You forgot the factor 1/2 when you calculated the energy.

 

- Except for the missing factor of 1/2 you correctly calculated the energy required to accelerate a car to a velocity (ignoring energy losses which in practice can probably be quite large) of 5 km per 5 minutes. I see no reason why you suddenly take that number and divide it by 1 second. I don't see what the result is supposed to represent. I think it's plain wrong. The average power required is the energy required/demanded divided by the time that the power will be applied. I'm not sure you need power at all.

 

- I've not read and checked the Snickers part; one problem at a time (in reality I'm just too lazy at the moment).

 

- Very nice description of what you tried. It's way too rare that people bother mentioning units and intermediate calculation steps .

[imp]

No. Acceleration and deceleration are ignored, as is friction. Not realistic, but interesting proposition nevertheless. imp

[heikediguore]

No. Acceleration and deceleration are ignored, as is friction. Not realistic, but interesting proposition nevertheless. imp

 

I thought I included acceleration by figuring out the change in energy. No? Ok, for simplicity's sake, so I don't have to calculate every force from friction to turbulence to gravity to nuclear decay, I'll say that the vehicle undergoes constant acceleration during its trip, such that the initial velocity is 0 m/s and the final velocity is 17 m/s.

 

I see no reason why you suddenly take that number and divide it by 1 second.

 

I was trying to find a formula for calculating total energy consumption, and I thought that I could do that by constructing power in terms of energy and time and then putting in a value for time.

 

I know I need to integrate something here. Is it:

 

integral[0,17](E(v)dv)

 

where E(v)=kinetic energy and v=velocity

 

so:

 

integral[0,17](1/2mv^2)dv

=[0,17](1/2mv^3)/3

=[0,17](mv^3)/6

=1000kg*(17 m/s)^3 - 0

=1000kg*4913 (m/s)^3

=4.913 * 10^6 kg (m/s)^3

 

Now I don't know what to do.

[timo]

I have doubts that you really need to integrate something. I am still not sure what you really want to calculate. I have the feeling you don't exactly know that yourself, so let me give you an offer of what you could calculate; maybe it comes close to what you try to do:

 

Assume you wanted to compare the energy required to accelerate a car from stop to a velocity of ~17 m/s (ignoring any resistance due to friction) with the "biological energy" of a Snickers (or alternatively calculate the number of Snickers whose bio energy equates it):

- The amount of energy required to accelerate the car equals the energy of the car after acceleration minus the energy of the car at rest.

- The amount of "bio energy" in a Snickers in SI units is obtained by converting the calories into joules.

- The number of Snickers whose "bio energy" equals the energy needed to accelerate the car equals the latter divided by the energy of a single Snickers.

[heikediguore]

Following some advice on another forum, I decided to take the following approach:

 

1. Assuming constant acceleration, with known initial and final velocities, use an equation of motion to find the acceleration.

 

2. Apply the acceleration to the work(force,distance) equation to find the energy requirement for such an acceleration over the given distance.

 

Here's what I got:

 

Equation of motion taken from wikipedia "Equations of Motion":

 

Vf^2= Vi^2 + 2ad

17 (m/s)^2 = 0 + 2a * 5000 m

289 (m/s)^2 = (10,000 m)a

 

so a=2.89 * 10^-2 m/s^2

 

W=Fd=mad

=1000 kg * 2.89 * 10^-2 m/s^2 * 5000 m

=1.45 * 10^5 kg (m/s)^2

=1.45 * 10^5 j

 

1 Snickers bar = 273 kcal

=2.73 * 10^5 cal

 

cal/4.184 = joules

(2.73*10^5 cal)/4.184

=6.52 * 10^4 j/snickers

 

so, using the Fd energy result:

(1.45 * 10^5 j)/(6.52 * 10^4 j/snickers)

=94.5 Snickers bars

[John Cuthber]

The question is poorly framed.

Imagine I start with a car at the top of a hill. It's a "physics question" car so there's no friction or air resistance.

I push the car and it rolls down the hill accelerating. Whan it gets to the bottom of the hill it runs up the other side of valley. After 5 mins and 5 Km (because I carfully chose a hill just steep enough to get the right journey time) it comes to rest at the top of the hill, across the valey from where it started.

Net energy consumed is nil.