A snowmobile climbing a hill...?

[KyleMe]

A snowmobile climbs a hill at velocity vup. The slope of the hill is a 1 ft rise for every 40 ft of distance. The resistive force of the snow is equal to 6.1 percent of the weight of the snowmobile. How fast will the snowmobile move downhill, assuming the engine delivers the same power? (Express the answer in mi/hr.)
Data: vup = 16.0 mi/hr.

I'm assuming conservation of energy is going to be a factor in this, but I'm not really even sure where to start and how to set up the problem.

 

[AsquareE]

we first have to calculate the power of the engine.
simply put P = F * vup
vup = 16 mi/hr
theta, the angle of the slope = arcsin1/40)
hence sin(theta) = 1/40
F = m*g*sin(theta) + resistive force of snow
F = m*g*1/40 + 6.1/100*m*g = m*g*(1/40+0.061) = m*0.84452
hence P = m*0.84452*16 N-mi/hr

now while going downhill, the force produced by the engine would be
F + m*g*sin(theta) = 0.061*m*g
F = m*g*(0.061 - 0.025) = m*g*0.036
F = m*0.35352
given that P is same
P = m*0.84452*16 = m*0.35352*vdwn
hence vdwn = 0.84452*16/0.35352 = 38.2222 mi/hr