A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of ki

Question

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.21 and the push imparts an initial speed of 4.0 m/s?

Answer 1

The force of friction = coefficient * normal force.
Normal force = mass * gravitational acceleration.
Substituting,
Ffr = coefficient * mass * gravity. We know the coefficient and gravity, so plugging in:
Ffr = 0.21*-9.8*m.

That's the sum of forces, which is equal to mass * acceleration. Guess what? The masses cancel out.
0.21*-9.8*m = ma
a = 0.21*-9.8 = -2.058 m/s^2
(acceleration is negative; the box is slowing down)

Using v^2 = (v0)^2 + 2*a*x to solve for x.
v^2, desired velocity, is going to be zero. You're solving for when it stops.
v0^2 is 4.0 m/s, given.
a = -2.058 m/s^2, you just calculated that.
-16 = -4.116x
x = 3.887 meters

Answer 2

Friction Force is in the direction of the Normal force which is
mgmu*dist = .5m4^2
cancel out terms
8=10*.21*dist=~4m