A 82.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 314.0 N. The floor is rough with a coefficient of friction of 0.30. What is the final speed of the crate, in m/s, after it has been pulled a distance of 13.0 m?

Answer 1

M*g = 82 * 9.8 = 803.6 N. = Wt. of crate. = Normal force, Fn.

Fk = u*Fn = 0.3 * 803.6 = 241 N. = Force
of kinetic friction.

F-Fk = M*a.
314-241 = 82*a.
82a = 73.
a = 0.890 m/s^2.

Vf^2 = Vo^2 + 2a*d.
Vo = 0.
Vf = ?.