# Work-Kinetic Energy question

1. Sep 18, 2011

### kiowaviator

This question is whooping me. I know I have the calculus part right I just dont know where to go from here?

A particle moving along the x axis is acted upon by a single force F = F0e^(–kx), where F0 and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of

F0/k
F0/e^(k)
kF0
1/2(kF0)^2
ke^(k)F0

So using the Work Kinetic Energy theorem
W=Kf-Ki since x=0 then Kf=W

W=integral(F = F0e^(–kx),x,0,xf)
taking F0 out, since its a constant, and integrating the exponential function I get
W=F0[-e^(-kxf)/k + 1/k]

now what????

2. Sep 18, 2011

### Spinnor

When you evaluate the integral one of the terms is zero,

exp(-x) goes to zero as x goes to infinity?

3. Sep 20, 2011

### kiowaviator

I did that, that's why the second term inside the brackets is 1/k, since e^(-k*0)=1.

4. Sep 20, 2011

### BruceW

You've pretty much done it now. So you now have 1/k inside the bracket and f0 outside the bracket, so just multiply them together to get the answer.

5. Sep 24, 2011

### kiowaviator

Ok, that makes sense. I was there I just couldn't see it. Thanks all.