Question

A 74 kg firefighter slides, from rest, 4.9 m down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor?

Answer 1

Answer:

Her speed is 9.8 meter per second

Explanation:

Newton's second law states that acceleration (a) is related with force (F) by:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

Here the only force acting on the firefighter is the weight F=mg so (1) is:

$mg=ma$

Solving for a:

$a=g$

Now with the acceleration we can use the Galileo's kinematic equation:

$Vf^{2}=Vo^{2}+2a\varDelta x$ (2)

With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

$Vf^{2}=2a\varDelta x$

Solving for Vf

$Vf=\sqrt{2g\varDelta x}=\sqrt{2(9.81)(4.9)}$

$Vf=9.8\frac{m}{s}$