Arightward force of 302 n is applied to a 28.6 kg crate to accelerate it across the floor. thye coefficient of friction between the crate and the floor is 0.750. determine the acceleration

We must all look for the fg, fn, Ff, and Fnet to get the acceleration.

1 Fg = 28.6 (-9.81)

= 0280.56 n

2 Fn = 280.56 N

3 Ff = μFn

= 0.750 * 280.56

Ff = 210.425 N

4. Fnet = Fa + Ff

= 302 – 210.42

Fnet = 91.576 N

5. F = ma

91.576 = 28.6a

a = 3.2 m/s

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6