The force on a particle is described by 9x^3 +5 at a point x along the -axis. Find the work done in moving the particle from the origin to x=5?
2007-04-28 16:28:37 · 2 answers · asked by toye7690 1 in Science & Mathematics ➔ Physics
The work done is the integral of force as a function of distance over the distance element, and is usually written in vector form as:
W = ∫F•dx,
where F and dx are vectors.
In this case, F and dx point along x, so your expression is scalar:
W = ∫Fdx
W = ∫[0,5](9x³+5)dx = (9/4)x^4+5x evaluated over [0,5]
W = (9/4)(5^4)+5(5) = (9/4)(625)+25
W = 1431 (I don't quote units because you didn't give me any.)
Good luck, work hard, and stay away from drgus.
2007-05-01 19:02:44 · answer #1 · answered by bridgejerk 2 · 0⤊ 0⤋
this would be an integral of the force from 0 to 5 of F*dx
the integral is
9/4*x^4+5*x
plugging in the limits
9*5^4/4+25
=1431 J
j
2007-04-30 17:54:23 · answer #2 · answered by odu83 7 · 0⤊ 0⤋