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gladu [14]
3 years ago
15

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 t

he frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor?
Physics
1 answer:
In-s [12.5K]3 years ago
6 0

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}

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