Question 436334
Since Work = Force*distance, we can use {{{F[net] = ma}}} where {{{F[net] = F - mg}}}, because the object is subjected to both F and its own weight. However there is a much faster way:


Note that the work done is equal to the total change in energy of the object. This is equal to the sum of the potential and kinetic energies gained, which is equal to {{{mgh + (1/2)mV^2}}} (mgh represents the potential energy and {{{(1/2)mV^2}}} is the kinetic energy of the object). Hence, this is the amount of work done.