# An engine can pull a 500 metric ton load up an inclined plane rising 1m in 100m of horizontal distance with a speed of 10m per second. The frictional force offered by the plane is 2000 N. What is the power of the engine? Given g = 9.8 meters per second squared.

Hello!

As I understand, "inclined plane rising 1 in 100" means that when travelling some distance along this inclined plane a body rises by 1/100 of that distance. In other words, the sine of an angle `alpha` of incline is `0.01.`

Also the metric ton is 1000 kg, so the mass m of a load is 500,000 kg.

Please look at the attached free-body diagram. There are 4 forces acting on a body: the gravity force `mg` downwards, the friction force `F_f` down along an incline, the reaction force `N` up perpendicularly to an incline and the traction force `F_T.`

Because a load moves with the constant speed `V` (no acceleration), the forces are balanced by Newton's Second law. Consider the x-axis parallel to an incline and consider the projections of the forces on it.

`F_T-F_f-mg*sin(alpha)=0`   (`N` is perpendicular to the x-axis and vanishes in this projection), so

`F_T=F_f+mg*sin(alpha).`

All the quantities at the right side are given.

Now find the power. It is by definition the work divided by the time `t.` Work is equal to `F_T*(V*t)` because the distance is `V*t` and the force and the displacement have the same direction. So the power is equal to

`(F_f+mg*sin(alpha))*V.`

In numbers it is (2000+500,000*9.8*0.01)*10=510,000 (watts) = 510 kilowatts.

The answer: the power of the engine is 510 KW.