A car on a hill - power of motor problem

[LippyKa16]

Homework Statement

A car is traveling up a slope. Given that a(t) is constant at 1m/s^2, find the expression of motor power in time. Assuming that the maximum power of the car is 215kW, for how long can the car maintain such an acceleration?

Mass of car (m) = 2000 kg
The hill has a constant inclination of 8 degrees (Θ = 8)
Initial velocity (u) = 90 km/h or 25 m/s
Friction coefficient due to wheel contact is 0.1
Air resistance (F[air]) = 0.85*v^2

Velocity = v, Time = t, Height = h, Force = F, Power = P
Energy = E, Kinetic Energy = KE, Gravitational Potential Energy = GPE

Homework Equations

P = ∑F * v
P = dE/dt
E = KE + GPE (no elastic potential energy)
GPE = 1/2*m*(h[final] - h[initial])
h[final] = s * sinΘ (s = distance)
s = 1/2(u+v)t (SUVAT equation)

The Attempt at a Solution

I drew a free body diagram of the system with a rotated axis so that the Normal force was the only force that was at an angle.

P[air] = F[air] * v * cosΘ * cos180
P[air] = -0.84*v^3

P[friction] = F[friction] * v * cosΘ * cos180 (F[friction] ~ 1942.91 N)
P[friction] = -1924.00*v

Assuming P[max] = P[motor]...
215,000 - 0.84*v^3 - 1924*v = dKE/dt + dGPE/dt
Maximum velocity means acceleration = 0
Maximum velocity also means KE = 0
Assuming h[initial] = 0

215,000 - 0.84*v^3 - 1924*v = d/dt (1/2*m*(1/2(u+v)t * sinΘ))
215,000 - 0.84*v^3 - 1924*v = d/dt(1000[(12.5t + 0.5vt)sinΘ])

I am stuck here because I think I am on the wrong track.
Any help is greatly appreciated.

 

[haruspex]

Friction coefficient due to wheel contact is 0.

I hope they mean rolling resistance. Static friction coefficient would be closer to 1, and is anyway irrelevant.

a rotated axis so that the Normal force was the only force that was at an angle.

Do you mean that gravity is the only one at an angle?

P[air] = F[air] * v * cosΘ * cos180

Why the the cos term here, and in the frictional force calculation? What is the angle between the velocity and the air resistance?

Maximum velocity means acceleration = 0

Yes, but the first part of the question asks for the power needed to achieve the given acceleration.

Maximum velocity also means KE = 0

You mean dKE/dt=0, yes?

d/dt (1/2*m*(1/2(u+v)t * sinθ))

What is this term? It looks a dKE/dt, not a dGPE/dt.

 

[LippyKa16]

I hope they mean rolling resistance. Static friction coefficient would be closer to 1, and is anyway irrelevant.

Do you mean that gravity is the only one at an angle?

Why the the cos term here, and in the frictional force calculation? What is the angle between the velocity and the air resistance?

Yes, but the first part of the question asks for the power needed to achieve the given acceleration.

You mean dKE/dt=0, yes?

What is this term? It looks a dKE/dt, not a dGPE/dt.

Sorry, yes I did mean the weight of the car was at an angle.

I was planning on adding the cosΘ term to all of the forces but that was unnecessary.

Yes sorry, I meant dKE/dt = 0 because maximum velocity is a constant.

The term d/dt (1/2*m*(1/2(u+v)t * sinΘ)) that I had was trying to incorporate the SUVAT formula for distance (s = 1/2(u+v)t) into the equation for final height (h = s * sinΘ)

 

[haruspex]

The term d/dt (1/2*m*(1/2(u+v)t * sinΘ)) that I had was trying to incorporate the SUVAT formula for distance (s = 1/2(u+v)t) into the equation for final height (h = s * sinΘ)

Ok, but there seems to be an extra 1/2 in there.
Anyway, it is much simpler to consider: if it is moving at speed v up a slope angle θ, how fast is it gaining height?

 

[LippyKa16]

Ok, but there seems to be an extra 1/2 in there.
Anyway, it is much simpler to consider: if it is moving at speed v up a slope angle θ, how fast is it gaining height?

So basically the s should be velocity?

 

[haruspex]

So basically the s should be velocity?

No, I did not say that. Simple question: if you are going up a slope θ at speed v how fast are you gaining height?

 

[LippyKa16]

No, I did not say that. Simple question: if you are going up a slope θ at speed v how fast are you gaining height?

You would be gaining height at v m/s right?

 

[haruspex]

You would be gaining height at v m/s right?

No, the speed v is up the slope, not vertical.

 

[LippyKa16]

No, the speed v is up the slope, not vertical.

So you would gain height at a speed of v*sinΘ? That would be the y component of velocity?

 

[haruspex]

So you would gain height at a speed of v*sinΘ? That would be the y component of velocity?

Right, so how much power does that represent?

 

[LippyKa16]

Right, so how much power does that represent?

Would that represent the v aspect of Power with regards to F*v for GPE?

 

[haruspex]

Would that represent the v aspect of Power with regards to F*v for GPE?

Not sure what you mean by "v aspect". It gives the power related to the gain of GPE. Yes, use the force times velocity formula to calculate it, but you must use the right force and the right velocity.

 

[LippyKa16]

Not sure what you mean by "v aspect". It gives the power related to the gain of GPE. Yes, use the force times velocity formula to calculate it, but you must use the right force and the right velocity.

Ok, thank you for your help so far.

 

[LippyKa16]

Not sure what you mean by "v aspect". It gives the power related to the gain of GPE. Yes, use the force times velocity formula to calculate it, but you must use the right force and the right velocity.

Would that Force be the force of the car and the velocity in v * sinθ is the unknown that I am trying to solve for? And would the equation for the gravitational potential energy be m*g*v*sinθ? I struggle with this stuff...

 

[haruspex]

would the equation for the gravitational potential energy be m*g*v*sinθ?

That's the rate of change of GPE, yes.

 

[LippyKa16]

That's the rate of change of GPE, yes.

Thank you for your help, I think I have solved it now.