Winch power for 950kg ore car on 30-degree incline

[cooltee13]

Homework Statement

A loaded ore car has a mass of 950kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0 degrees above the horizontal. The car accelerates uniformly to a speed of 2.2 m/s in 12.0 seconds and then continues at a constant speed.

a) what power must the winch motor provide when the car is moving at constant speed?

Homework Equations

Well, Power=Force x Velocity
Force = T

The Attempt at a Solution

Well to find the Tension in the rope, I found the component of G that was opposite the Tension and set it to -Fs. Then I set -Fs + T = 0 => Fs = T. Looking at The triangle under the cart I used sin(30)= Fs/g => Fs= gsin(30) which = 4.9 N

Since power = F*V I got P= (4.9)(2.2) = 10.78 kW. The back of the book says its 10.2 kW though. I am just wondering where I went wrong

 

[kamerling]

Well to find the Tension in the rope, I found the component of G that was opposite the Tension and set it to -Fs. Then I set -Fs + T = 0 => Fs = T. Looking at The triangle under the cart I used sin(30)= Fs/g => Fs= gsin(30) which = 4.9 N

I can't understand what you do with Fs and T. You do not use the mass of the cart. 4.9 N * 2.2 m/s will give you 10.78 W and not 10.78 kW

 

[cooltee13]

Oh ok, well then care to give me some advice on how to solve the problem then?

 

[kamerling]

Oh ok, well then care to give me some advice on how to solve the problem then?

The Idea of splitting the force of gravity in components parallel and perpendicular to the ground was ok. As was power = force * speed