# Find work done given time, acceleration, and mass

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?

## The Attempt at a Solution

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hi daltomagne! (try using the X2 icon just above the Reply box )
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

I am a little unsure as how to find d though. would it be one of the kinematic equations?
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 vf, and then apply the work-energy theorem ( work done = change in mechanical energy).

so i'm thinking v=vi+at and that gives me vi=-12.98m/s
and
xf=xi+vit+1/2at2?
but that gives me a value for x=-38.3 m?

where's the contradiction? one figure is speed, the other is distance, they should both give you the same work. 