Calculating Power Required for a Tank to Travel Uphill on an Incline

[Jonno]

Homework Statement

A Tank of mass 80 metric tons is travvelling at a uniform speed of 54Km/hr on a level terrian. It then starts traveling uphill on an incline of 1 in 10 (sine slope). Claculate the extra power required from the engine in megawatts to maintain the smae speed on the incline.

m = 80,000 kg
v = 15 m/s

Homework Equations

Kinetic Energy = $$\frac{1}{2}mv^2$$

Potential Energy = $$mgh$$

The Attempt at a Solution

You see, this is my problem. I'm at a loss on how to start I think I might be reading into this to0 much.

Is the extra power in MW required impossible to work out unless a distance traveled uphill is given in which case the following would be used:

Work = $$Fs\cos \phi$$

In which case how can I work the Force propelling hte tank in a horizontal direction?

The next part of the question is When it has traveled 100m up the incline the driver stops and has a drinks and so on. . . . . . . (I'm happy with this section of the question). Is this just the authors question writing or am I just reading too much into the original problem?

Thanks in advance Jonno.

 

[Doc Al]

Is the extra power in MW required impossible to work out unless a distance traveled uphill is given...

You don't need the distance you need the rate at which height is gained, which you can figure out given the speed and the angle of the incline. All you need to figure is the additional power needed to go uphill (against gravity).

 

[Jonno]

Thanks Doc Al,

I see where you're coming from.

So if the tank gains 1 metre in height for every 10m traveled in a second gains height at a rate of 1.5m/s

so using E = mgh I can work out the extra energy in joules and convert to Power (J/s) From then?

Thanks again.

 

[BigBadBart]

if you substitute h [m] by hdot [m/s] (with the dot on top, meaning it's the time-derivative) then E also becomes a time derivative. That's what you want. E [J] becomes P [W] (J/s=W)
so hdot is 1.5 m/s yes yes!

 

[Astronuc]

Going back to $$Fs\cos\phi$$ - change s to v, and use sin for changing elevation, as opposed to cos for horizontal displacement.