Question

A person doing a chin-up weighs 700 n exclusive of her arms. during the first 29.0 cm of the lift, each arm exerts an upward force of 355 n on the torso. if the upward movement starts from rest, what is the person's speed at this point?

Answer 1

`F_(g)` = 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

`F_(net)` = Net force acting on the torso = ?

net force on the torso is given as

`F_(net)` = F -
`F_(g)`

`F_(net)` = 710 - 700

`F_(net)` = 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

`F_(net)` d = (0.5) m v²

inserting the values

10 x 0.29 = (0.5) (71.4) v²

v = 0.285 m/s