An 88-kg fireman slides 5.6 m down a fire pole. He holds the pole, which exerts

Question

An 88-kg fireman slides 5.6 m down a fire pole. He holds the pole, which exerts a 500-N steady resistive force on the fireman. At the bottom he slows to a stop in 0.46 m by bending his knees.

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

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

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

Determine the acceleration of the fireman while stopping.

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

Explanation / Answer

1. Mg of fireman acting downward opposed by resisitive force of 500 N

using laws of motion

F = ma

88*9.81 - 500 = 88 * a

a = 4.128 m/s^2

the acceleration of the fireman while sliding down the pole is 4.128 m/s^2

2,)

using laws of motion,

Vf^2 = Vi^2 + 2*a*S { Vf is velocity at bottom, Vi is initial velocity i.e 0 , S is displacement and a is acceleration)

Vf^2 = 0 + 2*4.12*5.6

Vf = 6.692 m/s

3.)again, using equn of motion

Vf = Vi + at

t = 6.792/4.12

= 1.648 sec

4.)again, usinig equn of motion

Vf^2 = Vi^2 + 2aS

0 = 6.692^2 + 2*a*0.46

a = -48.677 m/s^2

5.) applying equn of motion

Vf = Vi +at

0 = 6.692 - 48.677*t

t = 0.1372 sec

time it takes for the fireman to stop after reaching the ground is 0.1372sec

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