A 1.75 kg box is pushed with a 8.35 N force across ground where yk = 0.267. What is the net force on the box?

Answer 1

Sum force on the X & Y direction

Variables

Force Applied = F

Weight of the Box = W

Force of friction = Fk

Now,

Sum Force X = F - Fk

Let Fk = μ_k * Fn

Therefore,

Sum Force X = F - μ_k * Fn

Now,

Sum Force Y = Fn - W

Therefore,

Fn = W = mg

Sub into Sum Force X equation

Sum Force X = F - μ_k * mg

Sum Force X = 8.35 - 0.267 * 1.75 * 9.81 = 3.766 N

Net Force = 3.766N