Question

An 88-kg fireman slides 5.1 m down a fire pole. He holds the pole, which exerts a 540-N steady resistive force on the fireman. At the bottom he slows to a stop in 0.43 m by bending his knees.

A) Determine the acceleration of the fireman while sliding down the pole.

B) Determine the velocity of the fireman just before reaching the grou

C) Determine the velocity of the fireman just before reaching the ground.

D) Determine the time it takes for the fireman to reach the ground.

E) Determine the acceleration of the fireman while stopping.

F) Determine the time it takes for the fireman to stop after reaching the ground.

Answer 1

Here,

mass of firearm , m = 88 Kg

resistive force , Fs = 540 N

a) let the acceleration of the fireman is a

Using second law of motion

mg - Fs = m * a

88 * 9.8 - 540 = 88 * a

a = 3.66 m/s^2

the acceleration of the fireman while sliding down the pole is 3.66 m/s^2

b)

d = 5.1 m

velocity of fireman just before reacing ground = sqrt(2 * a * d)

velocity of fireman just before reacing ground = sqrt(2 * 5.1 * 3.66 )

velocity of fireman just before reacing ground = 6.11 m/s

c)

d = 5.1 m

velocity of fireman just before reacing ground = sqrt(2 * a * d)

velocity of fireman just before reacing ground = sqrt(2 * 5.1 * 3.66 )

velocity of fireman just before reacing ground = 6.11 m/s

d)

let the time taken is t

using first equation of motion

v = u = a * t

6.11 = 3.66 * t

t = 1.67 s

the time taken to reach the ground is 1.67 s

e)

let the acceleration while stopping is a

using third equation of motion

0 - 6.11^2 = 2 * 0.43 * a

a = -43.4 m/s^2

the acceleration of the fireman while stopping is -43.4 m/s^2