Let F be a constant unit force that is parallel to the vector (-1,0,1) in xyz-space. What is the work done by F on a particle that moves along the path given by (t,t^2,t^3) between time t=0 and time t=1.

a) -1/4

b) -1/(4root(2))

c) 0

d) root(2)

e) 3root(2)

I did it wrong but my method was to take the projection of (t,t^2,t^3) on (-1,0,1)

Then integrating from 0 to 1. My answer was -1/4 (which is wrong its supposed to be 0) how did you guys do it?

## Question47 GR0568

After looking at it again you basically took the integral from 0 to 1 of the dot product of (-1,0,1) and the rate of change of (t,t^2,t^3).

Where the force on the particle is (1,2t,3t^2) since they want the force only parallel to (-1,0,1) its magnitude is 3t^2-1 at which point you simply integrate from 0 to 1.

Thanks for the help, I made two mistakes in my write up of this problem. Firstly I did not take the rate of change as the force I took the position vector itself and I did not take the magnitude of the projection vector.

Work is the dot product of force and displacement. Force is constant in this problem, but displacement is not.

So if r(t) is position, then total work is the limit of the sum of dot product of force and infitely small change in position i.e., integral over the curve of Fdr (lazy of me to write dot product like this!). Then just push around the symbols to get the answer.

### Re: Question47 GR0568

why are these physics questions like these

### Re: Question47 GR0568

I find this question weird, as it requires some knowledge of physics. Luckily I am a physics minor...

You can avoid all those nasty integrals (okay, not so bad, but still unnecessary) by knowing that the work done will be the difference in potential, and force is the gradient of potential, so taking U(x, y, z) = z - x to be the potential, U(1, 1, 1) - U(0, 0, 0) = 0.

The actual curve <t, t^2, t^3> is meaningless.

You can avoid all those nasty integrals (okay, not so bad, but still unnecessary) by knowing that the work done will be the difference in potential, and force is the gradient of potential, so taking U(x, y, z) = z - x to be the potential, U(1, 1, 1) - U(0, 0, 0) = 0.

The actual curve <t, t^2, t^3> is meaningless.