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Responding to an alarm, a 765-N firefighter slides down a pole to the ground floor, 3.5 m below. The firefighter starts at rest and lands with a speed of 3.8 m/s . Find the average force exerted on the firefighter by the pole.

Respuesta :

Answer:

Explanation:

Let acceleration of fall be a .

v² = u² + 2as

v = 3.8 m /s

u = 0

s = 3.5 m

3.8² = 0 + 2 x a x 3.5

a = 2.06 m /s²

Since this acceleration is less than g , an upward force is acting on the firefighter in the form of friction . Let this force be F . Let mass of the firefighter be m .

m = 765 / 9.8

= 78.06 kg

mg - F = ma

765 - F = 78.06 x 2.06

765 - F = 160.8

F = 604.2 N .

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