A40 n crate rests on a rough horizontal floor. a 12 n horizontal force is then applied to it. if the coecients of friction are μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:

Fr = 20 (N)

Explanation: See atached file ( free body diagram)

As for Newton´s low

∑ Fy  =  0

-mg + N = 0        ⇒  - 40 + N = 0       ⇒ N  = 40 [Newtons]

by definition :  Fr = μs * N        ⇒  Fr = 0,5 * 40      ⇒  Fr = 20 (N)

∑ Fx  =  0   body is at rest

Fe - Fr = 0

Fr > Fe

Fr > 12 (N)  the body is at rest

any one of the statements could be an accurate comparison. but the statement that would define the correct choice and seal the deal is not even on the list:

"the visibilities of vector-x and vector-y in the drawing provided with the question are identical and nonexistent."