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Work problem

  • Thread starter cyspope
  • Start date
  • #1
27
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Homework Statement


67. How fast must a cyclist climb a 6 degrees hill to maintain a power output of 0.25 horse power? Neglect work done by friction and assume the mass of cyclist plus bicycle is 70 kg.

A. 2.6 m/s

Homework Equations


W = mgh = Fdsin[tex]\theta[/tex]= [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]


The Attempt at a Solution


(0.25 hp * 735.5 W * 3600 J) = 70 * 9.8 * h * sin(6) = [tex]\frac{1}{2}[/tex]*70v[tex]^{2}[/tex]

661950=71.707h =35v[tex]^{2}[/tex]

I don't get how the answer should be only 2.6 m/s.
 
Last edited:

Answers and Replies

  • #2
54
0
I think your equations may be confusing you.

For instance what is h? h or (d) should be the distance that the cyclist is traveling.
so h=d sin (6) make sense?

now [tex] W = -\Delta U[/tex] so if you call your initial position h=0 your final potential is mgh

now can you use d to solve for velocity?
 
  • #3
27
0
but how do I find the velocity to maintain 0.25 hp?

661950 = 35v[tex]^{2}[/tex]
v[tex]^{2}[/tex] = 18912
v = 137.524 m/s

this is not what I should get. I need to get 2.6 m/s.
 
  • #4
1,860
0
This may help you out

[tex]P=\frac{dW}{dt}=\frac{F\cdot ds}{dt} = F \cdot v[/tex]

So you should figure out the force he needs to cancel out gravity, and from there get the velocity. Be careful of units, and it should be a straight shot.
 

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