A machine carries a 7.0 kg package from an initial position of at t = 0 to a final position of at t = 14.0 s. The constant force applied by the machine on the package is . For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package.

a) 189.2 J

b) 13.5 W

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:

We know the initial position:

and the final position:

So, the displacement is the difference between the two positions:

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:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

the components of a vector are the parts of the vectors that are to each other.

b perpendicular