A firefighter of mass 120 kg slides down a vertical pole with an acceleration of 4 m/s per second. What is the friction force that acts on the firefighters?

In this case, the friction force is the force that is acting against gravity.

Use F=M*a
If g=9.8m/s^2 and the firefighter's acceleration is 4m/s^2 then (9.8-4)m/s^2 is the deceleration suppied by the firefighter's hands (or legs, whatever part of the body he/she is using to decelerate). So the force is
F=M*(9.8-4)m/s^2 where M is the firefighter's mass.

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  1. 480kgm/s^2

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  2. Thank you

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