A person doing a chin-up weighs 700 N, disregarding the weight of the arms. During the first 24.0 cm of the lift, each arm exerts an upward force of 358 N on the torso. If the upward movement starts from rest, what is the person's speed at this point?
m/s
2007-12-18 06:05:19 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
F=ma
Force - weight =ma
(358x2)-700 =(700/9.8)a
a=0.224 m/s²
v²=2as+u²
v²=2(0.224)(0.24)+(0)
v=0.3279024245 m/s
2007-12-18 06:15:30 · answer #1 · answered by h_y_m_128 2 · 0⤊ 1⤋
Draw a FBD of the person
Sum of the forces = m*a
We know from F = m*a that m = F/a or W/g in this case
W/g * a = 2*F - W
Solve for a
a = g*(2*F)/W - g
a = g*((2*F)/W - 1)
Then use kinematics for the 2nd part
vf^2 = vi^2 + 2*a*d
vf = sqrt(2*g*((2*F)/W - 1)*.24)
Plug and chug
***
I got vf = 0.11 m/s
2007-12-18 06:15:12 · answer #2 · answered by civil_av8r 7 · 0⤊ 2⤋
F = ma so a = F / m;
The force and mass are known so solve for the acceleration m. Then use the given distance and that acceleration in
2 a d = v²
and solve for the final velocity v.
2007-12-18 06:11:02 · answer #3 · answered by jgoulden 7 · 0⤊ 2⤋
v=0.3279024245 m/s is correct.
2007-12-18 06:19:22 · answer #4 · answered by Geo-Star 2 · 0⤊ 1⤋