Question

Firefighters need to go from their resting quarters on the second floor to their equipment, 10 feet below, in a hurry.

(A) How fast would an 85 kg firefighter be moving at the bottom if he just jumped from the second floor to the first?
(B) What would be his kinetic energy when he got there?
(C) Now using a pole to slide down, the firefighter arrives at the bottom with a velocity of 1.5 m/s. How much work is done by the friction of the firefighter holding the pole?
(D) How great is the force of friction, based on your answer to C.

Answer 1

a) t = 0.622 s , v = 7.73 m / s , b) K = 2539 J , c) -2443.4 J , d) fr = 803.75 N

Step-by-step explanation:

A) when the fireman from the second floor is in a free fall motion the time to arrive, we can calculate it with kinematics

y = v t + ½ g t²

When you jump your initial speed is zero

y = ½ g t²

t = √2y / g

Let's reduce the magnitude to the SI system

y = 10 feet (0.3048 / 1 foot) = 3,048 m

t = √ (2 3,048 /9.8)

t = 0.622 s

The speed at this point is

v² = v₀² + 2 g y = 0 + 2g y

v = √ (2 9.8 3.048)

v = 7.73 m / s

B) The kinetic energy is

K = ½ m v²

K = ½ 85 7.73²

K = 2539 J

C) The relationship between work and kinetic energy is

W = ΔK =
`K_(f)` - K₀

The initial kinetic energy is zero because when its high speed is zero.

`K_(f)` = ½ m v²

`K_(f)`= ½ 85 1.5²

`K_(f)` = 95.6 J

W2 = 95.6 - 0 = 95.6 J

The variation of work in free fall less with friction

ΔW = W2 -W

Δw = 95.6 -2539

ΔW = -2443.4 J

This is the work done by the friction forces

C)

W2 = fr d cos θ

The angle is 180º because the friction opposes the movement

W2 = - fr d = -2443.4

fr = 2443.4 / d = 2443.4 / 3,048

fr = 803.75 N