Work is given by #W=vecFDeltarcos(theta)#, where #vecF# 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=vecFDeltay#
The force that must be exerted to lift the weight is its physical weight, given by #vecF_g=mg#. We are given #3kg# as a mass quantity, and #g# is the free-fall acceleration, #9.8m/s^2#.
#vecF=(3kg)(9.8m/s^2)=29.4N#
Therefore, a force of #29.4N# was exerted on the weight over a vertical distance of #1m#, and
#W=29.4N*1m#
#W=29.4J ~~24J#
Note that this is the work done by the lift, not the net work done on the weight. The net work done on the weight is #0J#, as the work done by the lift is equal and opposite the work done by gravity (#-24J#).
