1. ## Work and Energy

You are pulling your sister on a sled to the top of a 17.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the hill. If you pull from the bottom to the top, what will the speed of the sled be when you reach the top?

The other way is to use the equations of motion. Resolve the gravitational force so it is acting along the hill. Now you can get the resultant force acting on the sled, if you also resolve the force you pull with. Then use F=ma to get acceleration, and use the equations of motion.
The two methods are only superficially different, because we must remember the energy equations are derived (classically) from the equations of motion. But it is easier to solve it with the energy equations, as, and now we're going off topic, but they are more general, and the reason is a bit advanced, but if you cary on with physics, into university, you'll find out why.

3. Originally Posted by Linnus
You are pulling your sister on a sled to the top of a 17.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the hill. If you pull from the bottom to the top, what will the speed of the sled be when you reach the top?
Draw a sketch of the situation. I've got the hill sloping up and to the right. The angle between the plane of the hill and the horizontal is 10.0 degrees. The hill is 17 m at its highest point., giving us a length along the slope of $\displaystyle \frac{17~m}{\sin(10.0^o)} = 97.8991~m$. We've got a sled on the hill, with a mass of 50.0 kg. There is a rope connected to the sled making a 45.0 degree angle with the slope of the hill. You are pulling with a force of 127 N along this direction.

Okay. You know already that this is a work-energy problem, so let's get going with some definitions. As typical with my problem solving style I'm going to set the 0 point level for the gravitational potential energy at the lowest point in the diagram: the bottom of the hill. Since positive is upward for GPE, I'm just going to take the +h direction as straight up.

$\displaystyle W_{nc} = \Delta E$

We have no friction in the problem and gravity is conservative. The applied force F is almost always non-conservative. So
$\displaystyle \vec{F} \cdot \vec{s} = \Delta KE + \Delta PE$

$\displaystyle \left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) + (mgh - mgh_0) = \vec{F} \cdot \vec{s}$

At the bottom of the hill (the initial point) we are starting at $\displaystyle h_0 = 0~m$ and from rest, $\displaystyle v_0 = 0~m/s$.

So
$\displaystyle \frac{1}{2}mv^2 + mgh = \vec{F} \cdot \vec{s}$

Okay, let's talk about that dot product now. The applied force is at 45 degrees above the slope of the hill, so the angle between the force and the displacement is 45 degrees. Thus
$\displaystyle \vec{F} \cdot \vec{s} = Fs \cdot \cos(45.0^o) = (127~N)(97.89991~m) \cdot \cos(45.0^o) = 8791.59~J$

So
$\displaystyle \frac{1}{2}mv^2 + mgh = 8791.59$

$\displaystyle \frac{1}{2}mv^2 = 8791.59 - mgh$

$\displaystyle v = \sqrt{2 \left ( \frac{8791.59}{m} - gh \right )}$

I get $\displaystyle v = 4.29693~m/s$.

Pretty paltry for working almost 100 m up the slope!

-Dan