Homework Help: Help with a Kinetic/ Potential Energy problem please

1. Jun 17, 2010

Legal

1. The problem statement, all variables and given/known data
You are pulling your sister on a sled to the top of a 18.0 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?

2. Relevant equations
I'm not sure about these, but these are the few I tried to use...only to no avail :(.
1) ▲ X = 18.0 M/ Sin 10 (Hypotenuse = Opposite/ Sin Ө)
2) Force of Gravity on Sled = m*g*Sin 10
3) Total Work = (Force x Cos Ө) ▲ X
4) W = ▲PE + ▲KE
W= mg(▲Y) + ▲KE
Then I figured once I solved for ▲KE, I could use the following to determine Final Velocity:
5) ▲KE= (1/2)m(v^2)
Or simply v = Square Root(2 x ▲KE)/m
Alternate Equation Used:
Total Work Done = ▲KE = (1/2)m*(Final Velocity^2) - (1/2)m*(Initial Velocity)

3. The attempt at a solution

I got my initial displacement to be 104 m using Equation #1. Using Equation #2, I got the Force of Gravity on the Sled to be -85 N. So I Added 127 N + -85 N to get a Net Force of 42 N. I then used Equation 3 and plugged in 42 N * Cos 0 (Because there is no angle between the force and displacement) * 104 m to get roughly 4370 J. Using Equation 4, I calculated the change in Potential Energy to be 8830 J (50kg * 9.81 * 18m). This is where I get stuck because I'm certain there can't be a negative change in Kinetic Energy because it is derived from squaring a number.

I then went back and tried to use the result I derived from Equation 3 (4370 J), and I applied this number to the Alternate Equation I provided. I set 4370 J = ▲KE and tried to solve for Final Velocity. I used 0 m/s for Initial Velocity, essentially dropping the second half of the equation to zero, making 4370 J = (1/2) * mass * (FV^2). Solving for the Final Velocity, I received my answer to be 13.2 m/s. This would be quite a feat for a girl to be pulling a 50 kg sled up an incline that fast.

I know I'm messing up somewhere because I have yet to incorporate the 45 degree angle that the Force of the Tension is being applied from. Any help would be greatly appreciated!

Sorry if my explanation was poorly mapped out...I'm still very new to all of these concepts.

2. Jun 17, 2010

collinsmark

Sounds good so far.
Also good.
Oooh, not not so good.

Both the force pulling the rope, and the force of gravity, are both vectors and need to be treated as such. If you are going to add the vectors together, you need to add their respective components together -- not the overall magnitudes. But I don't recommend adding the vectors together for this problem. There is a much easier way that involves conservation of energy.
Ugh. Not following ya there. :uhh: Perhaps a free body diagram would be a good place to start.

Let's forget about the '42 N' figure, since we've already discussed why that's not correct, and go back to the original problem.

There is a 127 N force that is at an angle of 45o with respect to the displacement vector. If you want to find the work done by the pulling force (which you do want to find), you'll need to find the component of that 127 N force that is parallel with the displacement vector.

(Hint: that's one of the first steps to solving this problem: the work done by the pulling force. The second step is calculating the difference in gravitational potential energy [but this can be a completely separate step from the first step -- and actually the order is unimportant since both steps are independent. You could calculated the gravity part first if you wanted to]. Once you have the total work done, and the change in graviational energy, do you know how you would use them together to find the leftover kinetic energy? )

[Edit: Oh, by the way Legal, welcome to Physics Forums!]

Last edited: Jun 17, 2010
3. Jun 22, 2010

Legal

Wow! Excellent analysis. I thank you for your assistance and I apologize that it took me so long to do so. I greatly appreciate your help.