What is the acceleration of a 68.0 kg crate that is pushed across the floor by a 425 N force, if the coefficient of kinetic friction between the box and floor is 0.500? Please tell me where I went wrong because there's a part of it that I'm confused about.

Here's what I did:
Normal Force = Force of gravity = mg
= 68.0 kg (9.81 m/s²)
= 667 N
Ff = μ FN
= (0.500) (667 N)
= 334 N

Fnet = (667 - 425 - 334) N
= - 92 N

Fnet = mass x acceleration
-92 N = 68.0 kg (a)
a = - 92 N/68.0 kg

Question

What is the acceleration of a 68.0 kg crate that is pushed across the floor by a 425 N force, if the coefficient of kinetic friction between the box and floor is 0.500? Please tell me where I went wrong because there's a part of it that I'm confused about.

Here's what I did:
  Normal Force = Force of gravity = mg
					      = 68.0 kg (9.81 m/s²)
					      = 667 N
         Ff = μ FN 
                    = (0.500) (667 N)
	 = 334 N

        Fnet = (667 - 425 - 334) N
	   = - 92 N 		

		Fnet = mass x acceleration
	            -92 N = 68.0 kg (a)
	            a = - 92 N/68.0 kg

anonymous · Answer

Is the frictional force equivalent to the applied force in this case or this solution correct?