Power of a bicycle accelerating up a slope.

In summary, a cyclist can generate up to 141.6 watts of power to go up a slope at a 15 degree angle and 30 meters long.
  • #1
Kraken_Head
5
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No template because originally posted in wrong thread
Hello everyone :alien:
I have been assigned to create a physics problem involving bicycles. I thought up the following, but I had questions regarding if it was properly made / worked out.

A cyclist is accelerating constantly up a slope at 1 m/s ^-2 The slope is at a 15 degree angle and is 30 m long. It takes the cyclist 1 minute to reach the end of the slope. The combined mass of the cyclist and the bicycle is 80 kg. How much power is needed to make this possible ? (Neglect air resistance and drag)

I was thinking of using the formula: F = mg sin θ + ma to calculate the force needed to go up the slope, and then incorporating the found value of F into the formula: P = work / time = ( F x d ) / t to find the amount power needed to complete this task. The working out looks like this:

P = ( ( mg sin θ + ma ) x d ) / t
P = ( (( 80 x 9.81 ) sin ( 15 ) + ( 80 x 1 )) 30 ) / 60
P = 141.6 W

Is this correct ? Are there any variables of e.g. vectors, forces, etc. that I have forgotten to include in this calculation? or am I over complicating things ?

Thank you very much !
 
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  • #2
How do you know that after 1 minute, the bicycle only covers 30 m? I get 1800 m (almost 2 km). The work is also calculated incorrectly. The work is equal to the change in potential energy plus the change in kinetic energy. Also, the power is not constant. The work divided by the time is equal to the average power.
 
  • #3
Chestermiller said:
How do you know that after 1 minute, the bicycle only covers 30 m? I get 1800 m (almost 2 km). The work is also calculated incorrectly. The work is equal to the change in potential energy plus the change in kinetic energy. Also, the power is not constant. The work divided by the time is equal to the average power.

Thank you for making me notice that ! At the time, I was just trying to make up some figures off the top of my head, but now I realize I had to use the formula:
d = 1/2 a t^2 to figure out the distance traveled at that acceleration.
I also appreciate the correction on the work done formula. If I'm correct, it should look like this now :

W = Δ Ep + Δ Ek
____________________________________________________________________________________________

ΔEp = mg Δh
= ( 80 ) ( 9.81 ) Δ( length of X = 465.9 m - see right angled triangle )
= 319915.8 J

vtimxrajzbuard4hsj78_jpg_150×200_pixels.png

SOH
X = sin(15) x 1800 = 465.9 m
___________________________________________________________________________________________

ΔEk = Ek f - Ek i
= (1/2 m (v final)^2) - (1/2 m (v initial)^2)
v initial = 0
v final = (v initial) + at = (0) + (1)(60) = 60 m.s^-1
= (1/2 (80) (60)^2) - (1/2 (80) (0)^2)
= 144000 J
___________________________________________________________________________________________
W = (319915.8) + (144000)
= 463,915 .8 J

Average Power = work / time


Av P = ( 463,915.8) / 60
= 7731.9 W
____________________________________________________________________________________

However, upon seeing the final answer I can't help but feel like I miscalculated something ( I suspect an error when calculating the change in kinetic energy, or maybe the scenario simply isn't realistic ) because the magnitude of the answer - I feel - is too big for a cyclist.

Thank you for your feedback !
 
  • #4
Kraken_Head said:
However, upon seeing the final answer I can't help but feel like I miscalculated something ( I suspect an error when calculating the change in kinetic energy, or maybe the scenario simply isn't realistic ) because the magnitude of the answer - I feel - is too big for a cyclist.

The scenario isn't realistic, cyclists don't cycle at over 200km/hr (60m/s) and can't accelerate at a constant rate. They are close enough to constant power device, like a car. A quick google says an elite cyclist can maintain ~500W over an hour, ~1,200W in short bursts.
 
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  • #5
billy_joule said:
The scenario isn't realistic, cyclists don't cycle at over 200km/hr (60m/s) and can't accelerate at a constant rate. They are close enough to constant power device, like a car. A quick google says an elite cyclist can maintain ~500W over an hour, ~1,200W in short bursts.
Thank you !
 
  • #6
Perhaps simplify the problem by making it constant velocity rather than constant acceleration? or would that be too easy?

Perhaps turn the problem on it's head. The max power a cyclist can generate is 300W, how fast can he go up the slope?

How about a problem involving a unicyclist and working out the angle he must lean forwards or something like that?
 
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  • #7
You can still use the problem as stated, but you have some input numbers that you can play with to make the problem more realistic: the angle, the time, the acceleration. Play around with these until you get a more realistic power value.
 
  • #8
CWatters said:
Perhaps simplify the problem by making it constant velocity rather than constant acceleration? or would that be too easy?

Perhaps turn the problem on it's head. The max power a cyclist can generate is 300W, how fast can he go up the slope?

How about a problem involving a unicyclist and working out the angle he must lean forwards or something like that?

Thats a very good idea! Thanks for helping me out.
 
  • #9
Chestermiller said:
You can still use the problem as stated, but you have some input numbers that you can play with to make the problem more realistic: the angle, the time, the acceleration. Play around with these until you get a more realistic power value.

Thank you! Ill toy around with the equation a bit
 

Related to Power of a bicycle accelerating up a slope.

1. What is the power output of a bicycle accelerating up a slope?

The power output of a bicycle accelerating up a slope depends on several factors, including the weight of the rider, the steepness of the slope, and the gear ratio of the bicycle. Generally, the power output can range from 100-250 watts for an average cyclist.

2. How does the slope of the hill affect the power needed to accelerate?

The steeper the slope, the more power is needed to accelerate a bicycle. This is because the force of gravity pulling the rider and bicycle downhill increases, requiring the rider to pedal harder to overcome this force and maintain a constant speed.

3. Does the weight of the rider impact the power needed to accelerate?

Yes, the weight of the rider does impact the power needed to accelerate a bicycle up a slope. A heavier rider will require more power to overcome the force of gravity and maintain a constant speed, while a lighter rider will require less power.

4. How does the gear ratio of a bicycle affect its power output on a slope?

The gear ratio of a bicycle can greatly impact its power output on a slope. A lower gear ratio (easier gear) allows the rider to pedal at a higher cadence (number of pedal revolutions per minute) but with less force, resulting in a lower power output. On the other hand, a higher gear ratio (harder gear) requires the rider to pedal with more force, resulting in a higher power output.

5. Can a bicycle's power output be increased by changing gear ratios?

Yes, changing gear ratios can increase a bicycle's power output on a slope. As mentioned before, a higher gear ratio requires the rider to pedal with more force, resulting in a higher power output. However, it is important to find the right balance between gear ratio and pedaling efficiency to avoid tiring out too quickly.

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