V^2 = Vo^2 + 2a*d.
0 = 2^2 + 2*a*2/3,
a = -3 m/s^2.
Fap-Fk = M*a.
0-Fk = -3M,
Fk = 3M. = Force of kinetic friction.
uk = Fk/Fn = Fk/Mg = 3M/10M = 0.30
Fn = Normal force.
A rectangular box of length 2/3m is initially traveling at 2 m/s with its entire length over a smooth (perfectly frictionless)surface .The box gradually moves onto a rough surface and stops the instant that its entire length is positioned within the rough region.
Determine the coefficient of kinetic friction.
Details and Assumptions:
g=10 m/s²
The pressure at the bottom of the box is always uniform over its area.
1 answer