A 20 Kg box is sliding across a horizontal table at 10m/s. If the coefficient of Friction is 0.5, what is the acceleration of the box?

By Newton's second law,

• the net vertical force on the box is

∑ F = n - mg = 0

where n = magnitude of the normal force, m = mass of the box, and g = acceleration due to gravity; then

n = mg = (20 kg) (9.8 m/s²) = 196 N

• the net horizontal force on the box is

∑ F = -f = -ma

where f = mag. of kinetic friction and a = acceleration of the box. Since

f = µn

where m is the coefficient of kinetic friction, so that

f = 0.5 (196 N) = 98 N

Then the box's acceleration is

-98 N = - (20 kg) a   →   a = (98 N)/(20 kg) = 4.9 m/s²

there are no correct statementd on the list of choices you provided.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6