Calculating Power Output for a Cyclist Coasting Down a 7.2 Degree Hill

[Whenderson1]

Homework Statement

A cyclist coasts down a 7.2 degree hill at a steady speed of 10 km/h. If the total mass of the bike and rider is 75 kg, what power output must the rider have to climb the hill at the same speed?

Homework Equations

Power = Energy/Time
Ek = 1/2mv^2 where m is the mass v is the velocity and Ek is the kinetic energy
Eg = mgh where m is the mass g is a constant value of 9.8 h is the height above the ground and Eg is the gravitational potential energy.

The Attempt at a Solution

Power = Energy/Time = (Ek + Eg)/Time = [1/2(75)(2.777)^2 + (75)(9.8)(2.7sin7.2)]/1 = 522 Watts

Apparently the answer is 511 Watts. Can someone please tell me where i went wrong?

 

[PeterO]

Homework Statement

A cyclist coasts down a 7.2 degree hill at a steady speed of 10 km/h. If the total mass of the bike and rider is 75 kg, what power output must the rider have to climb the hill at the same speed?

Homework Equations

Power = Energy/Time
Ek = 1/2mv^2 where m is the mass v is the velocity and Ek is the kinetic energy
Eg = mgh where m is the mass g is a constant value of 9.8 h is the height above the ground and Eg is the gravitational potential energy.

The Attempt at a Solution

Power = Energy/Time = (Ek + Eg)/Time = [1/2(75)(2.777)^2 + (75)(9.8)(2.7sin7.2)]/1 = 522 Watts

Apparently the answer is 511 Watts. Can someone please tell me where i went wrong?

I don't like Power = Energy/time. Power is the rate of expending energy → a change in energy not the energy itself

I would be calculating using power = Force x velocity.

On the way down the hill, clearly the component of weight "driving" the cyclist down the hill is matched by friction - so acceleration is zero - the cyclist travels at constant speed.

On the way up the hill, the cyclist will have to create a driving force to over come each of those - his driving force has to match both friction and weight component.

Force x speed = power.