Calculating Power Output to Climb a 7.0 Degree Hill

[FossilFew]

A bicyclist coasts down a 7.0 degree hill at a steady speed of 5.0 m/s. Assuming a total mass of 75kg (bicycle plus rider), what must be the cyclist's power output to climb the same hill at the same speed?



I get 90N * 5.0 m/s = 450W



The book says 9.0 * 10^2 W

It looks like I'm at half the value of the correct answer. Any help is greatly appreciated!

 

[Locrian]

Just glancing at it, I got the same answer as you. Here we are using P=F*v, and that there is no net force on the bike. Maybe someone else can point out the problem here.

 

[FossilFew]

I think I have it. It's 450W *2 due to Frictional Force AND Force in the X direction. Thanks for your help.

 

[ammu]

please could you explain how to do this problem?

 

[NateTG]

please could you explain how to do this problem?

In addition to gravity, friction is doing work on the bicyclist. You need to account for the friction when determining the power necessary for him to climb the slope.

 

[ammu]

thanks for the reply,
actually i got the answer as 450 W but the right answer is 900W. i don't know why i can't get this answer! please help

 

[kevinyhong]

The amount of work in order to get the bike to go up the hill is equal to the amount of gravitational potential energy in addition to the kinetic energy needed to move at the same speed. that is why you get double the answer you've been getting: the change from potential energy to kinetic energy when coasting "down" the hill gives your velocity. you have to double it in order to maintain your kinetic energy as well as recover your potential energy that you lost from coasting down the hill.

 

[charliebrown9]

Sorry, i don't know anything about it.

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