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An 8.0-newton wooden block slides across a horizontal wooden floor at constant velocity. what is the magnitude of the force of kinetic friction between the block and the floor? (1) 2.4 n (3) 8.0 n (2) 3.4 n (4) 27 n

Verified Answer :

MissPhiladelphia

For the answer to the question above , the coefficient of kinetic friction(Ff) = 2.4N because: 
μ= .30 
Fg= 8 N 
m= Fg/g -> m= 8N / 9.81m/s^2= 80kg 
Fn = Fg 
Ff=μFn -> .308N = 2.4N
I hope my answer helped you. have a nice day!

skyluke89

Answer:

(1) 2.4 n

Explanation:

The force of kinetic friction between the block and the floor is given by:

\(F=\mu N\)

where

\(\mu\) is the coefficient of kinetic friction

N is the normal reaction of the floor against the block

For a horizontal surface (as in this case), the normal reaction is equal to the weight of the block, so \(N=8.0 N\). For wood-wood contact, the coefficient of kinetic friction is approximately \(\mu=0.3\). Therefore, the force of kinetic friction between the block and the floor is:

\(F=(0.3)(8.0 N)=2.4 N\)

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