Answer to Question #255761 in Mechanics | Relativity for xxxxx

Question

An 80-kg firefighter slides down a brass pole 8.0 m high in a fire station while exerting a frictional force of 600 N on the pole in the time of?

Answer 1

$ma=mg-F_{fr}$

$mx''(t)=mg-F_{fr}$

$x=(g-F_{fr}/m)t^2/2+c_1t+c_2$

if $v_0=0$ then $c_1=0,c_2=0$

So:

$t=\sqrt{\frac{2\cdot8}{g-F_{fr}/m}}=\sqrt{\frac{16}{9.8-600/80}}=2.6$ s