# Simple work and variable position question

1. Mar 11, 2012

### iJamJL

1. The problem statement, all variables and given/known data
A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by:

x = 0.484t^3 - 33.6t

Find W, the work done by this force during the first 1.49 s.

2. Relevant equations
W=F*s
W=mgs
Integration

3. The attempt at a solution
I just wanted to know whether I'm solving this correctly because I only have one more chance to input an answer in my online homework system (as those of you who have been helping me probably know by now.

Anyhow, this is what I've done.

W=Fs
F=mg
W=mgs
**Because the position is variable but the mass and gravity are constant, we integrate the formula given, and we get:

W=mg*∫x.dx= (.121t^4) - 16.8(t^2), from 0s to 1.49s

We come to:

W=mg*(-36.7)
W=6.26*9.81*(-36.7)= (-2253.8) J

Did I do this properly?

2. Mar 12, 2012

### ehild

The problem does not say that the particle is subjected to gravity. F=ma, and the acceleration is second derivative of position. Find the force from the given x(t) function.

ehild