force down = 9.81*12
max friction force = 9.81 * 12 * mu
forward force - friction force = m a
15 - 9.81*12 * mu = 12 (.2)
41. A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student uses a force of 15 N, what is the coefficient of kinetic friction of the floor?
Okay, so I know that the net force is 15 and the box weighs 12 kg. I think I am supposed to multiply but I am not sure if I need an angle
1 answer
1 answer