Work and Energy (snowmobile up- vs. downhill)

[Slickepot]

Taken from An introduction to mechanics - Kleppner, Kolenkow.
Problem 4.17

Homework Statement

A snowmobile climbs a hill at 15 mi/hr. The hill has a grade of 1 ft rise for every 40 ft. The resistive force due to the snow is 5 percent of the vehicle's weight. How fast will the snowmobile move downhill, assuming its engine delivers the same power?

Homework Equations

frictional force f=μF_n

P=dW/dt = F dx/dt = Fv

The Attempt at a Solution

angle of hill , tan θ = 1/40 , θ ≈ 0.025
Uphill forces : frictional force f = 0.05mg cos(θ)
and also mg sin(θ)

P = mg(0.05cos(θ) + sin(θ))v

Downhill forces : same frictional force, but now subtract for the gravitational pull in other direction. P is the same so,
v_down = P/F = \frac{mg(0.05cos(θ) + sin(θ))v}{mg(0.05cos(θ) - sin(θ)} ≈ 31 mi/hr

Answer says 45 mi/h in the book. So what have I missed, .. thanks for any suggestions.

 

[LawrenceC]

First thing, answer in book is correct.

"hill forces : frictional force f = 0.05mg cos(θ)"

above is not correct.

Hint: Determine the power consumed going up the hill in terms of weight. You have the speed, you have the rate of ascent, and you have the power consumed due to friction if you correct the above expression.

The only impediment going down hill is the friction. You have 'extra' power assisting you because of the rate of change of potential energy which can be expressed in terms of velocity and the 1 ft drop for every 40 ft of travel down the hill.

 

[Slickepot]

Oh ok, thanks for your reply.
An example of how I misinterpreted the given information. Thought the 5 percent of its weight was somehow 5 percent of its weight against the normal to the hill.