Responding to an alarm, a 764-N firefighter slides down a pole to the ground floor, 3.4 m below. The firefighter starts at rest and lands with a speed of 3.6 m/s .

Question

Find the average force exerted on the firefighter by the pole.

Express your answer to two significant figures and include the appropriate units.

Answer 1

First let's calculate the potential energy the firefighter has at the top of the pole. The potential energy is PE = mgh = 764 N x 3.4 m = 2598 Joules.

Now let's determine the mass of the firefighter, M = 764 N / 9.8 ms^-2 = 78.0 kg.

His kinetic energy at the bottom of the pole is

1/2 x mass x velocity² = 1/2 x 78.0kg x 3.6² m²s^-2 = 505 J.

The difference between the potential energy and the kinetic energy is the amount of energy lost from friction on the pole = 2093 J.

As work(energy) is force x distance, we can write F x d = 2093 J. As d = 3.4 m,

F - 2093/3.4 = 616 N which is the average force by the pole on the firefighter.

Please check the math.