Posted by **Shannon** on Tuesday, November 16, 2010 at 12:15am.

Starting from rest, a 98-kg firefighter slides down a fire pole. The average frictional force everted on him by the pole has a magnitude of 720 N, and his speed at the bottom of the pole is 3.8 m/s. How far did he slide down the pole?

- Physics -
**Nick**, Wednesday, October 22, 2014 at 8:05pm
You need to make a delta Y chart...

Y= unknown (?)

Vi= 0m/s (because he is starting from rest)

Vf=3.8m/s (bottom of pole)

a= found by using a=Fnet/mass equation below=7.34m/s

t= not needed

a=Fnet (720)/mass (98kg)= 7.34 m/s

To find how far he slid down the pole, you must use the Kinematic Equation that doesn't include time (Vf^2=Vi^2+2a(Y))--> 3.8^2=0^2+(2x7.34)(Y)-->14.44/14.68=14.68(Y)/14.68--> The pole 0.98 meters (3.2 feet)

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