Physics: 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?

Starting from rest, a 88-kg firefighter slides down

awesome267

09.01.2022

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

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