Given: #50 kg# sack lifted vertically a distance of #2 m#, and a #25 kg# sack lifted vertically a distance of #4 m#

Find the mechanical work done in each case.

#"Work" = F * d# provided the force acts in the opposite direction as the object's motion. Where #"Force" (F)# is measured in Newtons (#N#) and distance #(d)# is measured in meters #(m#)

The units of #"Work"# is the #"Joule" (J)#

#1 J = 1 N*m = (1 kg*m^2)/s^2#; #" "1 N = (1 kg*m)/s^2#

The weight of each sack exerts a force due to gravity: #W = m * g#, where #m = "mass" (kg)# and #g = 9.8 m/s^2#

#F_1 = 50 * (9.8) = 475 (kg*m)/s^2 = 475 N#

#F_2 = 25 * (9.8) = 245 (kg*m)/s^2 = 245 N#

#W_1 = 475 * 2 = 950 N*m = 950 J#

#W_2 = 245 * 4 = 980 N*m = 980 J#

The #25 kg# sack lifted #4 m# requires more energy.