Maximizing Cyclist Speed: Solving the Energy Question in a Race

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Homework Help Overview

The discussion revolves around a cyclist's speed during a race, focusing on the relationship between power, energy conversion, and kinetic energy. The original poster presents a scenario involving the cyclist's initial speed, power output, and the mass of the cyclist and bicycle, seeking to determine the final speed after a specific time interval.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore calculations involving energy input and kinetic energy, questioning the correct application of formulas. There are discussions about the initial kinetic energy and how it relates to the final speed. Some participants express confusion over the calculations and seek clarification on specific points.

Discussion Status

The conversation includes various attempts to clarify the calculations and approaches to finding the final speed. Some participants provide guidance on the correct use of kinetic energy formulas, while others express uncertainty about their results. There is no clear consensus, but multiple interpretations and methods are being explored.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. There are indications of confusion regarding the definitions and applications of kinetic energy and velocity changes.

neededthings
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A cyclist is competing in a race and decides to pass some fellow cyclists who are getting tired. The initial speed of the cyclist is 21 km/h and he has 3.0 * 10^2 W of power. If the bicycle is able to convert 92% of input energy into kinetic energy, how fast will cyclist be traveling after 4.0 s. The combined mass of the cyclist and the bicycle is 78 kg.

The answer is 7.9 m/s. But I don't know how.
 
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energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
KE= 1/2m( v)^2
root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
21km/h = 21000m/60x60s= 5.83 m/s (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s
 
useful energy =1.104x10^3 ... right
then change in kinetic energy will be equal to useful energy given in 4 seconds...
the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
so we get
change in kinetic energy = useful energy given..
mx(Vfinal2-Vinitial2)/2
 
I don't understand your solution. Can you please Clarify
 
neededthings said:
energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
KE= 1/2m( v)^2
root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
21km/h = 21000m/60x60s= 5.83 m/s (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s

Whoa, back it up a bit. You've got the energy input correct. What's your initial kinetic energy? Add the energy input to that to get the final kinetic energy. Use that to get a final velocity.
 
Kishlay said:
useful energy =1.104x10^3 ... right
then change in kinetic energy will be equal to useful energy given in 4 seconds...
the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
so we get
change in kinetic energy = useful energy given..
mx(Vfinal2-Vinitial2)/2

initial kinetic energy + useful energy given= final kinetic energy
 
hope this will make you clear
 
I get 21 which is not the correct? Please Help
 
  • #10
never mind i got it.

Thanks a lot
 
  • #11
he he he ...
if you have really got it then it is fine, if not we are always to help you...!
 

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