# Mechanics Problem - Power, Finding The Resistance and Acceleration

1. Apr 14, 2013

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
P=Fv
F=ma

3. The attempt at a solution
Part A: R = resistance to motion
$$P=Fv \\ 25000=F20 \\ F=\frac{25000}{20}=1250N \\ mgsin(arcsin(\frac{1}{15})) + R = 1250 \\ 784+R=1250 \\ R = 1250-784=466N \\$$

Part B:
$$F=ma \\ 1250-466=1200a \\ 784=1200a \\ a=\frac{784}{1200}=0.653ms^{-2}$$

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 :).

2. Apr 14, 2013

### tiny-tim

Your A looks ok to me, and your B is ok if the force is the same.

However, I suspect it's the power that's the same … otherwise why would they ask for the initial acceleration?

3. Apr 14, 2013

Thanks :)

Sorry I am confused now lol, Yeah the power output of the car remains constant I think, so my part B is wrong?

4. Apr 14, 2013

### tiny-tim

In part B, you've used the same F (1250) …

don't you need to use the same wattage?

5. Apr 14, 2013

Sorry if I am missing something obvious :shy:. Where can I factor the wattage in to part B? To find the acceleration I am using the force provided by the engine which is calculated using the wattage (25000/20) minus the resistance force. F=ma cant work if I used the wattage (25000) instead could it?

Thanks,

6. Apr 14, 2013

### haruspex

Since it asks for initial acceleration, the speed can be taken to be the same (as long as we assume the hill levels out smoothly). So if the power is the same then the thrust is also the same.
If the hill levels out suddenly it gets rather complicated. There's a trajectory to consider, the elasticity of the tyres, conservation of momentum...