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andrey2020 [161]
3 years ago
14

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

2 . What is the friction force that acts on him? Answer in units of N
Physics
1 answer:
natima [27]3 years ago
4 0

Answer:

The force of friction that acts on him is

F_k=567N

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

F=m*a

F=F_w-F_k

m*a=F_w-F_k

F_w=81kg*10m/s^2=810N

Sole to Fk

81kg*3m/s^2=810N-F_k

F_k=810N-243N

F_k=567N

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