Question

Starting from rest, a 94-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a magnitude of 750 N, and his speed at the bottom of the pole is 3.1 m/s. How far did he slide down the pole

Answer 1

Given that,

Mass of a firefighter, m = 94 kg

Frictional force, F = 750 N

Speed at the bottom of the pole is 3.1 m/s

It is required to find the distance he slide down the pole.

Using Work-energy theorem,

initial potential energy = final kinetic energy + work done

$mgh=W+\dfrac{1}{2}mv^2$

$mgh=Fh+\dfrac{1}{2}mv^2$

Solving for h, such that :

$h(mg-F)=\dfrac{1}{2}mv^2\\\\h=\dfrac{1}{2(mg-F)}mv^2\\\\h=\dfrac{1}{2(94(9.8)-750)}\times 94\times (3.1)^2\\\\h=2.63\ m$

So, he will slide 2.63 m down the pole.