# Non-Conservative Forces - Help, Please

1. Oct 24, 2009

### ksdeponte

Hello! I am having a lot of trouble with the following problem and would be very grateful for any help that you could offer me!

1. The problem statement, all variables and given/known data

There is a constant 125 N frictional force that opposes the motion of a soapbox car as it goes down a hill. The mass of the car is 100.0 kg and the angle of the hill is 55degrees, measured from the negative x axis. Its initial speed is 0.0 m/s and the final speed has to be 12.5 m/s in order for the car to make the jump successfully. The goal of the problem is to find the vertical height above the ground needed to have a final velocity of 12.5 m/s.
The problem also gives a hint to solve the problem, determine a way to express the distance the car travels in terms of its vertical height.

F = 125
m = 100.0 kg
theta = 55degrees, measured from the negative x axis (?), 180degrees for frictional force
vi = 0.0 m/s
vf = 12.5 m/s
delta x = ? m
h = ? m

2. Relevant equations

I've worked on this thing for hours and have no idea what to do.
I know that I will have to use the formula for non-conservative work, and that the final PE and initial KE are not needed. Also, Wnc will have to be in the form of (F cos theta)delta x, and that theta will be 180 since the work done by friction is opposing the car's displacement.

(F cos theta)delta x = KEfinal - PEinitial
= (1/2mvf^2) - (mgh)

But how would you determine delta x in terms of the vertical height, h? I've tried a multitude of things, but never get the right answer.

3. The attempt at a solution

Katrina DePonte

2. Oct 24, 2009

### cepheid

Staff Emeritus
Hello Katrina, welcome to PF!

Because of the frictional (non-conservative force), energy is not conserved, and instead of:

initial potential energy = final kinetic energy

initial potential energy - work done by friction = final kinetic energy

or,

initial potential energy = final kinetic energy + work done by friction,

The kinetic energy is known, and the other two things depend upon h. Your job is to find the value of h that satisfies the equation.

It looks like this is sort of what your equation is trying to be. However you have to be careful. The friction force points along the ramp, which is not the same thing as the x direction. The distance travelled along the ramp is related to the starting height by basic trigonometry. Hint: the distance along the ramp the hypotenuse of a right angle triangle, and the height is the vertical side of that triangle. Which trigonometric ratio is the ratio these two sides of a right triangle?

3. Oct 24, 2009

### willem2

This is simple trigonometry. draw a triangle with delta x and h as sides.

4. Oct 24, 2009

### ksdeponte

So, the work done by friction would not be (F cos 180)delta x, but (F sin 55)delta x?
which would then make delta x = h/sin 55.

PEi = KEf + (F sin 55)delta x
mgh = 1/2mvf^2 + (F sin 55)(h/sin 55)
100.0(9.8)h = 1/2(100.0)(12.5^2) + 125h
980h = 7812.5 + 125h
855h = 7812.5
h = 9.14 m

.. Is this the correct approach then?

5. Oct 24, 2009

### willem2

You have delta x in the direction of the slope, right?
delta x = h/sin 55 is correct, but the work done by friction is still (F cos 180)delta x.