# Homework Help: Energy and Power problem -- help please

1. Jul 12, 2016

### Traycee John

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

2. Relevant equations
K.E = 1/2mv2
GPE = mgh

3. 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!

2. Jul 12, 2016

### Staff: Mentor

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?

3. Jul 12, 2016

### Traycee John

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?

4. Jul 12, 2016

### Staff: Mentor

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?

5. Jul 12, 2016

### Traycee John

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 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:
Fnet=(750*9.8)+800
P=Fv, hence, v=P/v
Therefore, 30000/8150=v
3.68m/s=v

6. Jul 12, 2016

### Staff: Mentor

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?

7. Jul 12, 2016

### Traycee John

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!

8. Jul 12, 2016

### Staff: Mentor

Much better !!

9. Jul 12, 2016

### Traycee John

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
Fnet=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!!!

10. Jul 12, 2016

### Traycee John

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
Fnet=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!!!

11. Jul 12, 2016

### Staff: Mentor

Very nice.

12. Jul 12, 2016

### Traycee John

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