Calculating Work Done on 7.1 kg Object

[Crush1986]

Homework Statement

A force of ⃑ = 3.5 î+2.7 ĵ N acts on a 7.1 kg object over a displacement of = −2.2 î+3.9 j
m in a time period of ∆t = 0.28 s. At what average rate is work being
done?

Homework Equations

W= ΔKE
Kinematic equations
F=Ma
W= Fd

The Attempt at a Solution

I have tried a few ways, I know I'm wrong on the first and not sure of the second
The first way I tried to find the magnitude of force and displacement. I got that the Force's magnitude was 4.42N and the displacement was 4.478m. I then used W= Fd to find the work (4.42N)(4.478m)= 19.793J to find the rate of work I divided by the time (.28s) giving me 70.7 W. This is WAY off from the multiple choice answers which are .4W, .76W, 1.5W, 2.8W, and 10W. I'm not sure what my failure in logic is here. Maybe the force isn't directly pointed the same way as the displacement as I'm assuming? Or maybe something to do with resulting the displacement and force first... I'd love to know.

Attempt two: This time I focused more on the change in kinetic energy equaling the Work approach. Now for this I assumed the car was at rest to begin with... Not sure if that is ok. So I used the magnitude of the force (4.42N) and the weight of the car (7.1kg) to find the acceleration by a= F/m, (.6225 m/s^2) I used the given time to calculate the final velocity by doing a*t and that gave me a final velocity of .1743 m/s. I then calculated KE by 1/2mv^2 and divided it by time to find the rate of work. This gave me .39W, which is close to answer A.

Is attempt two correct? Or am I off...

I thank anyone for their time!

 

[Crush1986]

Ok, upon some further thinking, I think I dot product the two vectors and divide by time? Fd/t. So it ends up being 10w.

 

[PhanthomJay]

Ok, upon some further thinking, I think I dot product the two vectors and divide by time? Fd/t. So it ends up being 10w.

yes. The x comp of the force can only do work in the x direction and the y comp of the force can only do work in the y direction. You can add those work values up and divide by the time to get the same result.