Toni and the Bike: Physics Exam Puzzle

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the acceleration of a bike and the calculation of net force and power output. The original poster describes a scenario where a bike accelerates at 1.5 m/s² over a distance of 25 meters, with a total mass of 70 kg.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F = ma) to calculate the net force acting on the bike. There is also an exploration of the power output calculation, with attempts to apply relevant formulas such as W = Fd and P = W/t.

Discussion Status

Some participants have confirmed the calculation of the net force as correct, while others are working through the power output calculation. There is a mix of correct and incorrect reasoning being explored, with guidance offered on the proper formulas to use for power calculations.

Contextual Notes

Participants are navigating through the problem with some uncertainty regarding the correct application of formulas and units. There is a focus on ensuring clarity in the steps taken to arrive at answers, particularly for the power output calculation.

griefcrack
Messages
4
Reaction score
0
I'm going through my end of the year exam papers, since I figured I should actually make an effort at school this year, well anyway I'm terrible at physics and this question stumped me. Heres a diagram ( I drew it in MS paint ha, I even included a helmet cause riding bikes without one ain't safe. )

http://img214.imageshack.us/img214/9627/lolbikemy1.gif

While traveling home, the bike accelerates at 1.5 ms-2 over a distance of 25m as shown in the diagram.

Toni and the bike have a mass of 70kg.

(j) Using the equation F = ma, calculate the net force acting on the bike when it is accelerating.

(k) If it takes 5 seconds to cover the distance of 25m, calculate the power output of the bike during this time.

I think the answer for (j) is F = 70 x 1.5, so this gives me 105 N
Anyway I can't remember how I'm suppose to do (k) and my feeble attempts at doing it, gave me 13125 W but I'm sure that's wrong.
 
Last edited by a moderator:
Physics news on Phys.org
griefcrack said:
I think the answer for (j) is F = 70 x 1.5, so this gives me 105 N
That's correct
Anyway I can't remember how I'm suppose to do (k) and my feeble attempts at doing it, gave me 13125 W but I'm sure that's wrong.
What did you do to get this answer? Please show your work, as we can't see whether you're wrong if we can't see what you've done!
 
Thanks for replying so quickly, the other formula ( I think ) would be P = E x T and W = F x d, so I'm assuming W = 105 x 25 giving me 2625. Then P = 2625 x 5 giving me 13125 W.
 
griefcrack said:
Thanks for replying so quickly, the other formula ( I think ) would be P = E x T and W = F x d, so I'm assuming W = 105 x 25 giving me 2625. Then P = 2625 x 5 giving me 13125 W.

Your equation W=Fd is correct, and so is your numerical value of 2625. However, power is the rateof doing work, so your second equation is P=W/t.
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 44 ·
2
Replies
44
Views
8K
  • · Replies 38 ·
2
Replies
38
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K