Question

Playing on a playground, John Toleno ( 98 kg) slides down a pole to the ground floor, 14 m below. Dr. Toleno starts at rest and lands with a speed of 3.5 m/s.

(a) Find the average force exerted on Dr. Toleno by the pole.
____________ N
(b) If the landing speed is half that in part (a), is the average force exerted on Dr. Toleno doubled?

yes or no ?

(c) Find the average force exerted on Dr. Toleno when the landing speed is 1.75 m/s.
__________ N

Answer 1

a] Work done by the pole = Initial potential energy - final kinetic energy

=> Fh = mgh - (1/2)mv²

=> F(14) = 98(9.8)(14) - (1/2)(98)(3.5)²

=> F = 917.525 N

this is the average force exerted by the pole.

b] if v' = v/2 = 1.75 m/s, the average force is not doubled since the work done by the pole is related to v² and not v.

c]

F(14) = 98(9.8)(14) - (1/2)(98)(1.75)²

=> F = 949.68 N.