A certain force moves a 4-kg box from rest to a velocity of 6 m/s in 3 seconds. If the coefficient of friction between the box and the floor is 0.05, what is the magnitude of the force?
mass = 4 kg
μ (I'm assuming it means kinetic friction?) = 0.05
time = 3 seconds
initial velocity = 0 m/s
final velocity = 6 m/s
gravity = 9.8 m/s squared
Frictional Force = μ(mg)
Net Force = (m)(a)
final velocity = (initial velocity)(acceleration x time)
The Attempt at a Solution
From these equations, I calculated the acceleration to be 2 m/s squared. This is where I get confused - if you find the net force with the equation mass x acceleration, then you get 8 N for a net force. I calculated the frictional force with the equation above, so it was like this: (0.05)(4)(9.8) and I got an answer of 1.96 N. The natural force on the box is found by multiplying mass times the force of gravity, which ends up being 39.2 N. Now, my question: do I subtract the frictional force of 1.96 N (if that is right) from the natural force of 39.2 N to get the answer I'm looking for? Does the net force even get used in this problem? We only did a problem to find the acceleration with a given force, but not the other way around. Thanks to anyone that can help me!