Question

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 20.0-m tall building. the bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the building. what is the minimum amount of work that the worker did in lifting the bucket?

Answer 1

Using the work equation 
W = Fnet * d 
and Fnet = Fw + Fa (weight + force to accelerate) 
Fnet = m g + m a = m (g + a) 
we have m and g but need a 
let's use the equation; 
Vf^2 = Vi^2 + 2 a d 
4.0^2 = 0 + 2(20.0) a 
a = 16 / 40.0 = 0.40 m/s/s 
Therefore;  
W = Fnet d = m(g+a) d = 20.0(9.81+0.40)(20.0) = 4084 J 
or 4.084×10^3  Joule or 4.084 kJ 

Answer 2

For the work-energy theorem, the work done by the worker must be equal to the variation of mechanical energy of the bucket. 
The variation of mechanical energy of the bucket is sum of the variation of potential energy (U) and kinetic energy (K):
$\Delta E= \Delta U+\Delta K$
Taking the ground as reference, the variation of potential energy of the bucket is:
$\Delta U=mg \Delta h=(20.0 kg)(9.81 m/s^2)(20.0 m)=3924 J$
while the variation of kinetic energy is simply equal to the final kinetic energy of the bucket, since initially it was at rest (so its initial kinetic energy was zero):
$\Delta K= \frac{1}{2}mv_f^2= \frac{1}{2}(20.0 kg)(4.0 m/s)^2=160 J$

So the minimum amount of work that must be done is equal to the variation of mechanical energy of the bucket:
$W=\Delta E=\Delta U+\Delta K=3924 J+160 J=4084 J$