How much work is done on a bicyclist and her bike when coasting up a hill?

  • Thread starter Thread starter PhyzicsOfHockey
  • Start date Start date
  • Tags Tags
    Bicycle Hill
Click For Summary
SUMMARY

The discussion centers on calculating the work done on a bicyclist and her bike while coasting up a hill. The bicyclist, weighing 60.0 kg, and the bicycle, weighing 7.60 kg, start at a speed of 7.10 m/s and reach a speed of 6.00 m/s at the top of the hill. The incorrect calculation of work done was derived from the kinetic energy formula, resulting in -487 J, which was identified as erroneous. The correct approach requires careful application of the kinetic energy change formula, KEf - KEi = W, ensuring accurate mass and velocity values are used.

PREREQUISITES
  • Understanding of kinetic energy (KE) calculations
  • Familiarity with the work-energy principle
  • Basic algebra for manipulating equations
  • Knowledge of units in physics (kg, m/s, J)
NEXT STEPS
  • Review the work-energy theorem in physics
  • Practice problems involving kinetic energy calculations
  • Learn about energy conservation in mechanical systems
  • Explore the implications of negative work in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of kinetic energy and work calculations.

PhyzicsOfHockey
Messages
41
Reaction score
0

Homework Statement



Coasting along at 7.10 m/s, a 60.0 kg bicyclist on a 7.60 kg bicycle encounters a small hill. If the speed of the bicyclist is 6.00 m/s at the top of the hill, how much work was done on the bicyclist and her bike?


Homework Equations



KEf-KEi=W
.5mvi^2-.5mvf^2
kg*m^2*s^-2
m=60kg+7.60kg
v=m/s


The Attempt at a Solution



I tried finding the loss in KE and having that be the work done. I get

.5mVf^2 -.5mVi^2
.5*67.6kg*36 - .5*67.6kg*50.41

I get -487 J and it tells me this is the wrong answer? What did I do wrong?
 
Last edited:
Physics news on Phys.org
Please write out the whole equations. Include the mass and the units.
 
I tried finding the loss in KE and having that be the work done. I get

.5m(Vf-Vi)^2 and I get -487 J and it tells me this is the wrong answer? What did I do wrong?

Your final expression is incorrect mathematically.
 
I relized i posted that wrong but the answer comes out to -487 the right way but that's the wrong answer. I edited and wrote everything out.
 

Similar threads

Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
932
  • · Replies 4 ·
Replies
4
Views
7K
Replies
3
Views
6K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K