find work done given time, acceleration, and mass

daltomagne

a box with a mass of 6.0 kg is accelerated from rest by a force across a floor at a rate of
2.2 m/s^2 for 5.9 s. Find the net work done on the box

m=6.0kg
a=2.2m/s^2
t=5.9 s
a=0 degrees

i know W=Fdcos(a)
and F=ma


so,
F=6.0kg*2.2m/s^2=13.2N
W=13.2N*d*cos0=13.2N*d

I am a little unsure as how to find d though. would it be one of the kinematic equations?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

tiny-tim

hi daltomagne!

(try using the X² icon just above the Reply box )
That's right.

You can either use one of the standard constant acceleration equations to find d, or you can use another of them to find v_f, and then apply the work-energy theorem ( work done = change in mechanical energy).

daltomagne

so i'm thinking v=v_i+at and that gives me v_i=-12.98m/s
and
x_f=x_i+v_it+1/2at²?
but that gives me a value for x=-38.3 m?

so something isn't adding up

tiny-tim

where's the contradiction?

one figure is speed, the other is distance, they should both give you the same work.