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?

 

[tiny-tim]

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 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.

 

[rwisz]

Yes, you can use one of the kinematic equations to find the distance (d) traveled by the box. The equation that relates distance, initial velocity, acceleration, and time is d = (1/2)*a*t^2. Plugging in the given values, we get d = (1/2)*(2.2 m/s^2)*(5.9 s)^2 = 38.89 m.

Now, we can plug this value back into the work equation to find the net work done on the box:

W = 13.2 N * 38.89 m = 513.648 J

Therefore, the net work done on the box is 513.648 Joules. This means that the force exerted on the box over the given time and distance resulted in a transfer of 513.648 Joules of energy to the box.