[tex]F_{g}[/tex] = force of gravity acting in down direction on the torso = 700 N
F = applied force in upward direction by two arms together on the torso = 2 x 355 = 710 N
[tex]F_{net}[/tex] = Net force acting on the torso = ?
net force on the torso is given as
[tex]F_{net}[/tex] = F - [tex]F_{g}[/tex]
[tex]F_{net}[/tex] = 710 - 700
[tex]F_{net}[/tex] = 10 N
d = distance through which torso is lifted = 29 cm = 0.29 m
W = work done by the net force
m = mass of torso = weight/g = 700/9.8 = 71.4 kg
v = speed gained = ?
using work-change in kinetic energy theorem
W = (0.5) m v²
[tex]F_{net}[/tex] d = (0.5) m v²
inserting the values
10 x 0.29 = (0.5) (71.4) v²
v = 0.285 m/s
