# Problem with Forces and Newton's Laws

1. Sep 10, 2006

### catenn

Hi, I have a problem on my homework for AP Physics. I have the answer from the back but need to figure out why I am not getting the right solution. There is also an angle in the problem and I don't know what I am supposed to use it for or in what equation. The problem is:

A 40.0kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 140 N in it. Assume that the wagon starts from rest at the bottom of the hill and neglect friction. How fast is the wagon going after moving 80.0m up the hill?

The answer for it is 7.9m/s but I was wondering if anyone could explain how and why, especially what to do with the 18.5 degree incline part. Thanks!

2. Sep 10, 2006

### Staff: Mentor

I'd guess that 140N is more than what is necessary to hold the wagon steady on the incline, so the excess force available goes into accelerating the wagon. Calculate how much force is needed to hold the wagon steady, and then use the excess force to tell you what the acceleration is. Then use that to calculate the velocity as a function of time.

3. Sep 10, 2006

### Tensaiga

umm i solved it:
Convention is that up the slope is positive, down the slope is negative.

Force parallel to the slope:
Fx = mg sin @
= (40)(9.8)(sin18.5)
= 124.38N
since this force is down the slope it's considered negative. (You are pulling it up the slope, that force can be considered the sliding force.)

we know that Fnet = Tensional + Fx (since no fricition)
= 140N + (-124.38N)
Acceleration = 15.62N / 40
A = 0.3905 m/s^2

Now you know it's starting form the rest, so it's initial velocity is zero,
use this $$v^2 = v_0^2 + 2 a \Delta x$$ formula to find the final velocity after traveling 80m.

Hope this helps.

4. Sep 10, 2006

### Israfil

The forces applying on the wagon are F=140 N (uphill), F_h (downhill), F_g (Gravity) and F_n (at right angle onto the uphill ground). Now you need to draw them and figure out F_h because F - F_h is the force that pulls up the wagon. You'll discover from your drawing the F_h = F_g*sin(18.5°) with F_g=m*g (m, mass of wagon, g=9.81m/s^2)
Then you have F_h and F - F_h = F_result.
Since F=m*a -> F_result = m_wagon*a gives you a, the accelaration of the wagon. Then its movement equation is s=1/2*a*t^2 with s=80m. This gives you t, the time the wagon need to travel 80m. By v=a*t you get the speed of the wagon after a certain time.
Then you have what you need.
Good luck :)

5. Sep 10, 2006

### Staff: Mentor

Please remember Tensaiga and Israfil, that we are not supposed to do the poster's work for them. We are supposed to offer suggestions and clear up confusions, but NOT do their work for them. Please be sure that you have read the homework forum guidelines -- they are very explicit about not giving out answers.

6. Sep 10, 2006

### Israfil

hey berkeman,

yes sorry, you're right... I was just waiting for someone to help me with the question I tried to figure out for hours without any luck, so I thought I could help answer some other questions meanwhile...

greets :)

7. Sep 10, 2006

### Tensaiga

ah i see, sorry, we were suppose to give out hints.

8. Sep 10, 2006

### catenn

Thank you so much Tensaiga and Israfil, your explanations cleared up a lot. I already had the answer from the back of the book and just wanted to know how to do the problem. I learn by seeing how something is done and working through it as I look at it. AP Physics is really hard, I hope it is something that eventually becomes easier as you get used to it.