Solving Power Output for 3400 KG Truck on 8° Incline

[AZhang]

A 3400 KG truck is driving up a road inclined at 8 degrees above the horizontal. If the resistive forces such as air drag and friction that act on the car add up to 1200 N, determine the power output necessary to keep the velocity constant at 20 m/s.

I don't really get how to set up my equations. Can someone help me?

 

[Delphi51]

Have you got a formula for finding power? If not, try wikipedia.

 

[AZhang]

I do, it's both Work/Time and Total Energy/Time, but I don't know how to use them in this scenario.

 

[Delphi51]

You need pick a convenient time OR distance to work with. Say you choose 1 second for the time. Then you must work out the distance along the incline that it goes in that time. And the height straight up that it goes. These numbers should allow you to calculate the energy used to overcome friction and the energy used to increase the height. Finally, P = energy/time should do the trick.

 

[rwisz]

To solve this problem, we can use the equation P = Fv, where P is power, F is force, and v is velocity.

First, we need to calculate the total resistive force acting on the truck, which is the sum of air drag and friction. This is given as 1200 N.

Next, we can use trigonometry to find the component of this force acting against the direction of motion, which is equal to the weight of the truck. This can be calculated as follows:

Weight of truck = mg = (3400 kg)(9.8 m/s^2) = 33320 N

Component of resistive force = 1200 N * sin(8°) = 1200 N * 0.139 = 167 N

Now, we can use this value to calculate the net force acting on the truck:

Net force = 33320 N - 167 N = 33153 N

Finally, we can plug this value into the power equation to find the power output required to maintain a constant velocity of 20 m/s:

Power = (33153 N)(20 m/s) = 663060 watts or 663 kW

Therefore, the truck would need a power output of 663 kW to maintain a constant velocity of 20 m/s while driving up an 8 degree incline with a total resistive force of 1200 N.