a) 189.2 J
b) 13.5 W
Step-by-step explanation:
a)
The work done by the force on the package is equal to the scalar product between force and displacement:

where F is the force and d is the displacement.
In this problem, the force is:
![F=(5.0)i+(6.0)j+(10.0)k[N]](https://img.qammunity.org/2021/formulas/physics/college/ixp1ymxt0ffihxv2syf5csf5204vpp624n.png)
We know the initial position:

and the final position:

So, the displacement is the difference between the two positions:
![d=d_f-d_i=(7.5-0.6)i+(15.0-0.75)j+(7.2-0.28)k=\\=6.9i+14.25j+6.92k[m]](https://img.qammunity.org/2021/formulas/physics/college/jfs0a5f92h3nbkx0cuh3snguy3kzpjnj56.png)
Therefore, the scalar product between force and displacement is:

b)
The average power of the machine is the rate of work done per unit time; it is given by

where
W is the work done
t is the time elapsed
In this problem, we have:
W = 189.2 J (work done)
t = 14.0 - 0 = 14.0 s (time elapsed)
Therefore, the average power of the machine is:
