How Do You Calculate Maximum Car Speed Against Resistance and Incline?

[Traycee John]

Homework Statement

A car of mass 750kg has a maximum power of 30kW and moves against a constant resistance to motion of 800N. Find the the maximum speed of the car:
i) on level
ii) Up an incline of S^-1 1/10 to the horizontal

Homework Equations

K.E = 1/2mv²
GPE = mgh

The Attempt at a Solution

I am not sure how to approach this problem as I am just learning about the topic of Energy and Power on my own for a course to get extra creds. I've looked in a couple texts and online to get an idea of how to approach these kind of equations but they all leave me confused. How would you approach this question? Thank you!

 

[Chestermiller]

You must have some thoughts on this. What are they? If the speed of the car is v, in terms of v, how much power is being exerted against the constant resistance?

 

[Traycee John]

You must have some thoughts on this. What are they? If the speed of the car is v, in terms of v, how much power is being exerted against the constant resistance?

Well, so far, what I think I got is the force the car uses to accelerate which is mg (750*9.8). Using the equation Power=F*v, I guess the velocity would be v=P/F?

 

[Chestermiller]

Well, so far, what I think I got is the force the car uses to accelerate which is mg (750*9.8). Using the equation Power=F*v, I guess the velocity would be v=P/F?

mg is the weight of the car. How is that related to its forward acceleration in part (i)? Is the kinetic energy of the car changing once it reaches maximum velocity? The rate of power consumed against the resistance is equal to the resistance force (800 N) times the velocity, right? If the car has reached maximum speed, how is maximum rate of power provided by the car related to the rate of power consumed against the resistance?

 

[Traycee John]

mg is the weight of the car. How is that related to its forward acceleration in part (i)? Is the kinetic energy of the car changing once it reaches maximum velocity? The rate of power consumed against the resistance is equal to the resistance force (800 N) times the velocity, right?

Yes, mg would be the gravitational force of the car. It's related to the forward acceleration as it more than the constant resistance of 800N. As a body on level ground, with the perpendicular forces in equilibrium, the total net force would be 800N+Mg which requires a power of 30kW for it's forward acceleration, I'm assuming? The Kinetic energy is not changing as the motion is constant. What would this mean for the net Kinetic energy? Energy is being lost, but I don't know how much.

If the car has reached maximum speed, how is maximum rate of power provided by the car related to the rate of power consumed against the resistance?

The maximum rate of power provided by the car would be lower than the rate of power consumed against the resistance, as the car needs more power to get to maximum velocity against the resistance? I'm not sure to to utilise the equation k=1/2mv²

This is what I attempted to do:
F_net=(750*9.8)+800
P=Fv, hence, v=P/v
Therefore, 30000/8150=v
3.68m/s=v

 

[Chestermiller]

Yes, mg would be the gravitational force of the car. It's related to the forward acceleration as it more than the constant resistance of 800N. As a body on level ground, with the perpendicular forces in equilibrium, the total net force would be 800N+Mg which requires a power of 30kW for it's forward acceleration, I'm assuming? The Kinetic energy is not changing as the motion is constant. What would this mean for the net Kinetic energy? Energy is being lost, but I don't know how much.

The 800 N is in the horizontal direction and the mg is in the vertical direction. They can't be included together. The work and power in the vertical direction are zero, because the displacement in the vertical direction is zero. At maximum velocity, the 30 kW are not being used to accelerate the car. It is just being used to maintain a constant speed against the resistance force.

The maximum rate of power provided by the car would be lower than the rate of power consumed against the resistance, as the car needs more power to get to maximum velocity against the resistance?

At the maximum velocity, the maximum rate of power provided is equal to the rate of power consumed against the resistance.

I'm not sure to to utilise the equation k=1/2mv²

You don't need to use it because the kinetic energy is constant at maximum velocity.

This is what I attempted to do:
F_net=(750*9.8)+800
P=Fv, hence, v=P/v
Therefore, 30000/8150=v
3.68m/s=v

This isn't correct. the 750*9.8 shouldn't be in there. Did you really think that the maximum velocity of the car was only 3.68 m/s?

 

[Traycee John]

The 800 N is in the horizontal direction and the mg is in the vertical direction. They can't be included together. The work and power in the vertical direction are zero, because the displacement in the vertical direction is zero. At maximum velocity, the 30 kW are not being used to accelerate the car. It is just being used to maintain a constant speed against the resistance force.
At the maximum velocity, the maximum rate of power provided is equal to the rate of power consumed against the resistance.

You don't need to use it because the kinetic energy is constant at maximum velocity.

This isn't correct. the 750*9.8 shouldn't be in there. Did you really think that the maximum velocity of the car was only 3.68 m/s?

Thank you very much for the insight! I understand what you're saying hahaa. So the equation needed for this problem is P=f/v?

Using the equation:
P=Fv
v=P/F
v=30000/800
v=37.5m/s

Is this correct? I'm sorry, but this is actually the first time I'm coming across this Topic and Physics in general!

 

[Chestermiller]

Thank you very much for the insight! I understand what you're saying hahaa. So the equation needed for this problem is P=f/v?

Using the equation:
P=Fv
v=P/F
v=30000/800
v=37.5m/s

Is this correct? I'm sorry, but this is actually the first time I'm coming across this Topic and Physics in general!

Much better !

 

[Traycee John]

Much better !

Yes, the original equations kind of threw me off because those are what were given on the sheet of paper I had.

Anyway, I attempted the second part:
ii)Up an incline of Sin-1 1/10 to the horizontal

In the horizontal direction we have the Constant force of Resistance of 800N and the force due to the incline (mgsin(thetha)).

mgsin(sin^-11/10)=735
F_net=735+800
=1535N
v=P/F
v=30000/1535
v=19.5m/s

Does this seem correct? Thanks for your help once again, it was much appreciated!

 

[Traycee John]

Much better !

Yes, the original equations kind of threw me off because those are what were given on the sheet of paper I had.

Anyway, I attempted the second part:
ii)Up an incline of Sin-1 1/10 to the horizontal

In the horizontal direction we have the Constant force of Resistance of 800N and the force due to the incline (mgsin(thetha)).

mgsin(sin^-11/10)=735
F_net=735+800
=1535N
v=P/F
v=30000/1535
v=19.5m/s

Does this seem correct? Thanks for your help once again, it was much appreciated!

 

[Chestermiller]

Yes, the original equations kind of threw me off because those are what were given on the sheet of paper I had.

Anyway, I attempted the second part:
ii)Up an incline of Sin-1 1/10 to the horizontal

In the horizontal direction we have the Constant force of Resistance of 800N and the force due to the incline (mgsin(thetha)).

mgsin(sin^-11/10)=735
F_net=735+800
=1535N
v=P/F
v=30000/1535
v=19.5m/s

Does this seem correct? Thanks for your help once again, it was much appreciated!

Very nice.

 

[Traycee John]

Very nice.

Have a great night, Sir. I think I am ready for my Midterm paper tomorrow! :)