Force, speed of truck on a 10.0 degree hill

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Valerie Prowse
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Homework Statement



A loaded truck has a mass of 3100 kg. The maximum speed it can maintain on a 5.0° hill is 80 km/h. What constant speed could the truck maintain on a hill with a slope of 10.0°? Assume the total force due to air resistance and friction is 700 N and that it does not vary with speed.

Homework Equations



∑F = ma

The Attempt at a Solution



I have an exam tomorrow and this is one of the practice questions. The answer is 44.8 km/h, but I have no idea how to get there. What I have worked out so far is the FBD of the truck, and:
since a = 0, for θ = 10
∑F = 0
F - Ffr - mg⋅sin10 = 0
F = Ffr + mg⋅sin10
F = 5975.4 N
and vθ=10 = ??

I am having trouble relating this back to a velocity. Also, since there is no variation in air resistance and friction, it would mean that:
for θ = 5
F = Ffr + mg⋅sin5
F = 3347.8 N
and vθ=5 = 22.2 m/s

I feel like I am almost there but I can't quite put the pieces together ... I've though about work and momentum, but those don't seem to fit with the idea of the question ...
Does anyone have an idea on where to go from here??
Thanks!
 
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Have you considered thinking in terms of the power the truck can deliver?
 
gneill said:
Have you considered thinking in terms of the power the truck can deliver?

I'm not sure what you mean by that..
 
Valerie Prowse said:
I'm not sure what you mean by that..
While the truck is climbing the 5° slope it is said to be doing the maximum speed that it can achieve on that slope. So it (or rather its engine) must be delivering the most power it can under the circumstances. Can you determine that power?
 
gneill said:
While the truck is climbing the 5° slope it is said to be doing the maximum speed that it can achieve on that slope. So it (or rather its engine) must be delivering the most power it can under the circumstances. Can you determine that power?

Ah, I see the problem. This question is from a previous exam, but my course did not cover this topic, hence why I had no idea what you meant by power. I went through my textbook and found the section on power, which was not assigned, but I will give it a quick look over and try this question again later.
Thank you for your help, though!