# Is this a constant or variable force?

1. Apr 6, 2009

### revolution200

If a vehicle is stopping its change in Kinetic energy is equal to the force times displacement or distance as it is traveling in a straight line and parallel to displacement.

If the force is F(net) = kinetic + drag

F(net) = umg + 1/2*A*rho*v^2

is this force constant or variable

If it is constant then the Kinetic energy (T) is

delta(T) = F(net) delta(s)

If it is variable then it is the formula for variable force

What I'm hoping to get is a differential equation of the form

dT/ds=F(net)

2. Apr 6, 2009

### Hootenanny

Staff Emeritus
The force is clearly a function of velocity and hence time and is therefore, non-constant.

What exactly are you trying to solve for?

3. Apr 6, 2009

### revolution200

I have solved it for distance.

dT/ds = B+CT where v^2=2T/m

solving I got

s = ln(B+CT) - ln(B) = ln(B+CT/B)

looking back through definitions I have noticed that I have used a formula for constant force when it isn't but I'm not sure

4. Apr 6, 2009

### Hootenanny

Staff Emeritus
I'm not sure why/how you're invoking energy here. And no, as I said above you cannot treat the force as constant.

A must more straightforward method would be to simply note the definition of force in terms of acceleration and solve the resulting ODE.

5. Apr 6, 2009

### revolution200

I have a vehicle stopping. I have already solved this issue and am writing the result up. I have descibed the stopping distance of the vehicle as a differential equation that describes the rate that the vehicle is losing energy. When the vehicle has stopped its Kinetic energy is zero.

The solution I have for this is

Where Work = change in Kinetic energy = Force * displacement

dT = F(net)ds

dT/ds = F(net)

where F(net) = umg + drag equation http://en.wikipedia.org/wiki/Air_resistance

for the v^2 term in the drag equation I substitute V^2 = 2T/m

so

F(net) = B + CT

the frictional force plus the drag force in terms of kinetic energy

I have also calculated a numerical solution using an Euler method that iterates down using the differential equation.

The problem I have had is do I get the same result for variable force as I do with constant force
i.e.
dT = F(net)ds

6. Apr 6, 2009

### uart

You write it as an ODE.

$$dT/ds = -b - cT$$

Which has the solution,

$$T = (T_0 + b/c) e^{-cs} - b/c \,\,\,\,\,$$ : valid while T>0

BTW. Have you done ODE's in maths yet?

7. Apr 6, 2009

### revolution200

yeah I've done one or two. I needed to solve it for s(max). The solution to the problem is how far the car travels not the kinetic energy of the vehicle.

from
dT/ds = b+cT

we get

dt/(b+cT) = ds

for

(b+cT)=x, dx = cT

therefore

T = 1/c(dx)

we now get

1/c (dx/x)=ds

integrating

integral b+cT0 - b(1/c){1/x} =integral 0 - smax{s}

we get

smax = 1/c{ln(b)-ln(b+cT0}

The constants are negative.

is there a way ogf using equation editor

I just need to know if I can get

dT = F(net)ds from the variable force if you can help