- #1
FaraDazed
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Homework Statement
A small car with a mass of 1200kg is driving up a hill with a slope of sin^-1(1/15) at a constant velocity of 20m/s. The power developed by the engine is 25000W.
Part A: find the resistance to motion.
Part B: At the top, the road becomes horizontal. Find the initial acceleration, assuming the resistance is unchanged.
Homework Equations
P=Fv
F=ma
The Attempt at a Solution
Part A: R = resistance to motion
[tex]
P=Fv \\
25000=F20 \\
F=\frac{25000}{20}=1250N \\
mgsin(arcsin(\frac{1}{15})) + R = 1250 \\
784+R=1250 \\
R = 1250-784=466N \\
[/tex]
Part B:
[tex]
F=ma \\
1250-466=1200a \\
784=1200a \\
a=\frac{784}{1200}=0.653ms^{-2}
[/tex]
Part B I am a bit unsure of, my mind tells me the net force is the starting force (1250) minus one lot of resistance (466), however two of my peers have had the total force to be 466. Also it depends on whether I got part A correct and I am not 100% on that either.
Any help appreciated :).