Mechanics Problem - Power, Finding The Resistance and Acceleration

[FaraDazed]

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
$$P=Fv \\<br /> 25000=F20 \\<br /> F=\frac{25000}{20}=1250N \\<br /> mgsin(arcsin(\frac{1}{15})) + R = 1250 \\<br /> 784+R=1250 \\<br /> R = 1250-784=466N \\$$

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

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 :).

 

[tiny-tim]

Hi FaraDazed!

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?

 

[FaraDazed]

Hi FaraDazed!

Your A looks ok to me

Thanks :)

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?

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

 

[tiny-tim]

Yeah the power output of the car remains constant I think, so my part B is wrong?

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

don't you need to use the same wattage?​

 

[FaraDazed]

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

don't you need to use the same wattage?​

Sorry if I am missing something obvious :shy:. Where can I factor the wattage into 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 can't work if I used the wattage (25000) instead could it?

Thanks,

 

[haruspex]

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?

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...