How Is Maximum Kinetic Energy Calculated for an Electric Bicycle with a Rider?

  • Thread starter Thread starter Dongorgon
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion focuses on calculating the maximum kinetic energy (KE) of an electric bicycle with a rider, considering an 80kg rider and a 30kg bicycle at a maximum speed of 6m/s. The initial calculation of KE using work done led to an incorrect value of 34,496,000J, which was deemed excessive. The correct approach involves using the formula KE = 1/2 mv^2, resulting in a KE of 1980J, which raised concerns about its sufficiency given the bicycle's ability to travel 32km. It was clarified that the maximum KE does not equate to the total energy required for the 32km journey, as the battery's energy is not solely converted to kinetic energy. The discussion emphasizes the importance of distinguishing relevant information in problem-solving.
Dongorgon
Messages
20
Reaction score
0

Homework Statement



An engineer was to design a bicycle that can be powered only by an electric motor. The bicycle can carry an 80kg rider at a maximum speed of 6m/s. Under these conditions the maximum distance the bicycle can cover before the battery needs to be recharged is 32km. The bicycle has a mass of 30kg. Calculate the maximum kinetic energy of the bicycle and rider when the rider is not pedalling.


The Attempt at a Solution


Assuming that no resistance force acts, what I've done is to consider the work done as equal to the change in kinetic energy over the entire displacement.
From this: WD=110x9.8x32000=34496000J
Hence, KEmax=34496000J
Which also is the entire stored energy of the battery.
Is this a correct way, and would this answer be sufficient? Thanks
 
Physics news on Phys.org
You seem to have involved g somehow, but there's no mention of vertical movement. What's your logic for that?
OTOH, you have not used the info re 6m/s, which does seem to be highly relevant.
(Are there more parts to the question?)
 
Yeah, through working out the KE via 1/2mv^2 it obtains a value of 1980J. But for some reason, this seems far too small considering this energy can carry the rider 32km also? Would I be correct in thinking this value for KE is also the total energy stored by the battery?
Also, there are no more parts, just that. Thanks for the fast reply.
 
Dongorgon said:
Yeah, through working out the KE via 1/2mv^2 it obtains a value of 1980J. But for some reason, this seems far too small considering this energy can carry the rider 32km also?
The max KE is not the energy required to carry the rider 32km. That might be the case if all the energy in the battery were converted to KE straight away and thereafter it's just carried along by its KE. The reason I asked if there are more parts is that the 32km does not seem to be relevant. But it's not unusual for irrelevant information to be provided as a test of whether the solver can figure out what is.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top