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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

answer///isobars;