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I have a bicycle weighing 8 kilos. An 10 kilometer ride averaging 20 km/h requires z kWh's of energy.

How much more energy would I spend if I added 2 kilos to my backpack?

Or turning it around: If I used the same amount of energy, how much slower would I get?

UPDATE: I'm asking out of curiosity, not out of some homework assignment. It's a long time since I've had those :)

I have friends arguing that a 2 kg lighter bike (e.g. a racer instead of a cyclo cross) will give me serious advantages because of the decreased weight, even though I'm an amateur when it comes to bicycling. Me, on the other hand, can't really see why this would matter all that much until I get seriously more fit. Anyway, I'm not familiar with the physics needed to calculate this.

If it helps, let's assume a 5 degree rise from 0-5 km in 10 km/h and a 5 degree descent from 5-10 km in 40 km/h.

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closed as off-topic by Dilaton, ja72, Waffle's Crazy Peanut, Emilio Pisanty, akhmeteli Aug 23 '13 at 17:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Dilaton, Waffle's Crazy Peanut, Emilio Pisanty, akhmeteli
If this question can be reworded to fit the rules in the help center, please edit the question.

Please note that Homework questions are supposed to supposed to show more effort. For more details, please see the Homework Policy. – centralcharge Aug 22 '13 at 12:06
This isn't answerable. It depends on how much you accelerate/decelerate during the bike ride, on the difference in altitude between start and finish and on the airodynamics of the bicyclist. The force needed for acceleration and increasing altitude goes linear with the mass, but air resistance doesn't depend on mass at all. – Volker Aug 22 '13 at 12:33
@Volker, I don't agree with you that this is not answerable. You could assume that air resistance will do the same work in both cases, so it will not affect the difference in work done. And acceleration/deceleration won't affect the work done, which only depends on mass and final velocity. There is still the problem of altitude, and frictional losses, but they can be approximated. – udiboy1209 Aug 22 '13 at 12:47
Yes, but then you need to know what fraction of the total energy spent went towards overcoming air resistance (and possibly other types of friction) out of the total energy. Without this additional information, no answer can be given. Take two extreme cases: You accelerate a little bit at first on a plane, go at constant velocity against the wind until the end. The 2 kilos will barely make a difference. However, if you're in the mountains where the air is thin but you go up a 30% slope, the increase in energy will be almost 25%. Maybe even more if you need to stop every kilometer – Volker Aug 22 '13 at 12:53
@neu242: Is this really homework or are you trying to solve a real world problem? – Volker Aug 22 '13 at 12:55

If you are doing this out of curiosity for a specific route: The easiest way to get a rough estimate is to look at the elevation profile of your route and assume that when going downhill you are expending the same amount of energy regardless of the weight (which might not be accurate), and base all your energy changes when going uphill. Your change in energy requirements on a specific uphill would be:

$$E_{change} = (m_{new}-m_{old})*g*h$$

If you know the average speed up the hill, your average power change during only that hill would be:

$$P_{change} = (m_{new}-m_{old})*g*V_{avg}$$

If you do this for all hills, and assume downhills and flats are the same power, you can average all your $P_{change}$ for all those sections and get your average $P_{change}$.

If your trip is mostly flat, my guess would be that your power savings would be almost unnoticeable, unless like somebody commented, you are just stopping and accelerating all the time while trying to keep the same average speed for both mass conditions.

If this is a homework question: I think the way to do it is like yankeefan11's answer here using $E = 1/2*m*v^2$. The distance comes into play when you want to calculate the change in power.

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Thanks! It's not a homework question, I'm just trying to understand how these physics work, and if those two kilos really matter much. I am not convinced that they do :) – neu242 Aug 27 '13 at 6:19
If you're thinking about how much "faster" you can go with a bike which is 2 kilos lighter, well the truth is not so much (even though 2 kilos difference might be a lot of money difference on a bike). Most important on a bike is that you're properly fitted to it; when properly fitted you will be able to put as much power TO the bike as any biker with a 10+grand set. Sure, not as much power might go to speed... but the way I see it, you're just training with weights ;) – Esteban Aug 27 '13 at 18:20
Thanks. That's my opinion as well :) – neu242 Aug 28 '13 at 7:11

We can use the fact that our energy will be given as $$E = \frac12mv^2$$ So assuming that you can neglect air resistance and that you keep the same speed, and everything is kept simple, the additional energy you need is: $$E_{new}-E_{old}$$ $$\frac 12m_{new}v^2-\frac 12 m_{old}v^2$$ $$\frac 12v^2(m_{new}-m_{old})$$

You know the difference in mass that you added, so you can just plug in your values.

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So it doesn't matter whether you are travelling 10 or 20km? – Bernhard Aug 22 '13 at 13:28
It does. Energy is given as $.5mv^2$, I stated that I assume the same speed is kept – yankeefan11 Aug 22 '13 at 13:31
Where is this distance in your equations? – Bernhard Aug 22 '13 at 14:01
I did not account for it... – yankeefan11 Aug 22 '13 at 14:25
This is just one form of energy. There are lots more for bicycle riding. – ja72 Aug 22 '13 at 16:39

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