Question

A 12 kg crate initially at rest is pushed across the floor. The coefficient of kinetic friction between the crate and the floor is 0.190. If the crate was pushed with a force of 90.0 N at an angle of 40° below the horizontal, what is the final speed of the crate after it has moved 2.3 m?

Answer 1

Ok so first you need to find the fnet,
Fnet = fx-Ff
Fx = F • sin
Ff = coefficient of friction • Fn
Fn = Fg + Fy
Fg = m • ag

So now we fill in all of this.

Fg = 12 • 9.81 = 117.72
Fn = 117.72 + 0 = 117.72 (fy is 0 because there isn’t an downward force)
Ff = 0.19 • 117.72 = 22.3668
Fx = 90 • sin 40 = 57.85
Fnet = 57.85-22.3668 = 35.4841

Now that we have the fnet we can find our acceleration.

F = m•a

We need to find a so we switch the F and a around

a = F/m
a = 35.4841/12 = 2.95m/s^2

Now that we have our acceleration we can use Vf^2 = Vi^2 + 2ad

Vf^2 = 0^2 + 2 (2.95) (2.3) = 13.57
Vf^2 = 13.57
Vf = 3.68 m/s

Now we have our final velocity.