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 790 N, and his speed at the bottom of the pole is 3.3 m/s. How far did he slide down the pole?

Answer 1

U(intial velocity)=0m/s Weight=88*g Friction force=790N V(final velocity)=3.3m/s
Weight-Frictionalforce=Mass*acceleration
880-790=88*a
a=90/88=1.022727m/s^2
V^2=U^2+2as
S(distance)=(3.3)^2-(0)^2/2*1.022727

Answer 2

Your formula worked!! Just remember gravity is 9.81m/s^2 not 10m/s^2. Thank you!