Question

Starting from rest, a 88-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a magnitude of 800 N, and his speed at the bottom of the pole is 3.2 m/s. How far did he slide down the pole?

Answer 1

Answer:h=7.22 m

Explanation:

Given

mass of firefighter $m=88 kg$

Frictional Force $F=800 N$

average speed at bottom $v=3.2 m/s$

Let h be the height of Pole

net force on Firefighter is

$F_{net}=88\times 9.8-800=62.4$

therefore net acceleration is

$a_{net}=\frac{62.4}{88}=0.709 m/s^2$

using $v^2-u^2=2ah$

here $v=3.2 m/s$

$u=0$

$(3.2)^2-0=2\times 0.709\times h$

$h=\frac{10.24}{1.418}$

$h=7.22 m$