An 82-kg fireman slides 5.2 m down a fire pole. He holds the pole, which exerts a 550-N steady resistive force on the fireman. At the bottom, he slows to a stop in 0.45 m by bending his knees.
Determine the acceleration of the fireman while stopping. Express your answer with the appropriate units. Enter a positive value if the acceleration is downward and negative value if the acceleration is upward.

Explanation:

Let the frictional upward force be F = 550 N

Net force on it = mg - F

82 X 9.8 - 550

= 803.6 - 550

253.6 N

Downward acceleration = 253.6 / 82

= 3.0927 m / s²

final velocity after slipping 5.2 m , v then

v² = 2 x a s

= 2 x  3.0927 x 5.2

v = 5.67 m /s

It stops in .45 m distance

v² = u² - 2as

0 = 5.67² - 2 a x .45

a = - 35.72 m / s²

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