Question

Responding to an alarm, a 96 kg fireman slides down a pole to the ground floor, 3.2 m below. The fireman starts at rest and lands with a speed of 3.6 m/s.

(a) Find the average force exerted on the fireman by the pole.
N

(b) If the landing speed is double that in part (a), is the average force exerted on the fireman halved?

yes or no   

Explain.

(c) Find the average force exerted on the fireman when the landing speed is 7.2 m/s.
  N

Answer 1

In this problem lets assume the fireman is holding onto the pole with the same grip force.

First draw a picture. Place an origin at the bottom of the pole. Then make a free body diagram of the fireman. Using our kinematic equations we can find the acceleration of the fireman.
one of our equations says that Vf^2 = Vi^2 +2*a*(Xf - Xi)
lets plug the known values into the equation.
Vf = 3.6 m/s Final Velocity
Vi = 0 Initial velocity
Xf = 0 m
Xi = 3.2 m
Plug in and get your acceleration. a= 2.025 m/s^2

Acceleration due to gravity (free fall) is 9.81 m/s^2.

So the difference in the firefighters acceleration and free fall 9.81 - 2.025 = 7.785.
Remember F = m*a. Use it: F = 96 * 7.785 = 747.36 N of force excerted on the fireman by the pole to retard his fall.

B)

No

explanation: as you can see in C) the answer is less than half.

C) Just change the numbers in the first equation. a' = 8.1 m/s^2

So the difference in the firefighters acceleration and free fall 9.81 - 8.1 = 1.71m/s^2

and F'=96 * 1.71 = 164.16N

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