Asked by AJ

A firefighter of mass 73 kg slides down a vertical pole with an acceleration of 5 m/s2. The acceleration of gravity is 10m/s2. What is the friction foce that acts on him? Answer in units of N
Answered by Damon
Fdown = m g
Fup = friction force

F = m a

m g - Fup = m a
so
Fup = m(g-a) = 73 (10 - 5) = 365 Newtons
Answered by Henry
Fg = m*g = 73 * 10 = 730 N. = Force of gravity.

Fg-Fk = m*a
730-Fk = 73 * 5 = 365 N.
-Fk = 365-730 = -365
Fk = 365 N. = Force of kinetic friction.

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