Work is given by #W=FDeltarcos(theta)#, where #F# is the applied force, #Deltar# is the displacement, and #theta# is the angle between the force and displacement vectors. Because the weight is moving upward and this is the direction that the force of the lift is applied, #theta=0^o#, #W# is positive, and #W=FDeltay#
The force that must be exerted to lift the weight is its physical weight, given by #F_g=mg#. We are given #6kg# as a mass quantity, and #g# is the free-fall acceleration, #9.8m/s^2#.
#F=(6kg)(9.8m/s^2)=58.8N#
Therefore, a force of #58.8N# was exerted on the weight over a vertical distance of #21m#, and
#W=58.8N*21m#
#W=1234.8J ~~1235J#
In scientific notation for significant figures, #W~~1*10^3J#.
Note that this is the work done by the lift, not the net work done on the weight.
