Question

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

Answer 1

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=μ*Fn -> .30*8N = 2.4N
I hope my answer helped you. have a nice day!

Answer 2

(1) 2.4 n

Step-by-step 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`