posted by Shannon
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?
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)