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

Step-by-step 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* 9.8-800=62.4`

therefore net acceleration is

`a_(net)=(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* 0.709* h`

`h=(10.24)/(1.418)`

`h=7.22 m`