work and power output-- Help please, don't know what Im doing?

Question

A helper lifts 35 bricks to the top of a 2.8 m wall. Each brick weighs 1300g.
how much work is done neglecting the weight of his own hands to lift all the bricks?

Is it w=mg------9.8*2.8=27.44m/s^2 or w=fXd=1300*35=45,500*2.8=127,400

It takes 3 mins to lift the bricks....what is his power output?

---1300g/1000=1.3
1.3*35=45.5
45.5*2.8=127.4/3s=42.5s
or 27.44/3s= 9.1?

a motor driven conveyor produces 3000W of useful output. It carries the bricks to the top of the wall. how long would the conveyor take to lift the blocks?

Answer 1

Let´s take a coherent system of units to work out

mass kg length meter and time second

In this system the weight of each brick is 1.300*9.8 Newton and the work done elevating 35 bricks is

W= 35*1.3*9.8*2.8 Joule =1,248.52 Joule

Power = work/time = 1248.52/180 =6.94 watt

if you use 3000watt as work/power = time

time = 1,248.52/3000=0.42 seconds

Answer 2

The force of the helper against gravity is f=m*g 1.3Kg*9.8m/s^2=12.749N per brick
the total work is 12.79n*2.8m*35=1249.36J.
The work is done in 3min=3*60s=180s
P=1249.36J/180s=6.94W
The power of the machine is 3000 so t=W/P=1249.36J/3000W=0.41s