Question

1. 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

Solution) h = 10 feet

1 foot = 0.305 m

h = 10×0.305 = 3.05 m

(A) m = 85 kg

V = ?

Applying conservation of energy

PE = KE

mgh = (1/2)(m)(V^2)

V^2 = 2gh

V = (2gh)^(1/2)

V = (2×9.8×3.05)^(1/2)

V = 7.73 m/s

(B) Kinetic energy , KE = ?

KE = (1/2)(m)(V^2)

KE = (1/2)(85)(7.73^2)

KE = 2539.49

KE = 2539.5 J

(C) V' = 1.5 m/s

Work done , W = ?

W = change in kinetic energy

W = (1/2)(m)(V'^2) - (1/2)(m)(V^2)

W = (1/2)(85)(1.5^2) - (1/2)(85)(7.73^2)

W = - 2443.87 J

(D) Friction force , f = ?

W = - 2443.87

- (f)(h) = - 2443.87

f = (2443.87)/(3.05)

f = 801.26 N