A steel box is being pushed along a wooden table top by an applied force of 120N forwards. If the normal force between the box and table is 150N, and it accelerates at 3.0 m/s^2 [F], what is the coefficient of kinetic friction?
Show your work please :)
Show your work please :)
-
We know that Fnet/mass = 3m/s²
We know 150N = mass*g => mass = 150/g = 15.3kg
(120-150*uk)/15.3kg = 3m/s² => 120N - 150*uk = 3m/s²*15.3kg = 45.9N
(120N-45.9N)/150N = uk = 0.4938 <----------------
We know 150N = mass*g => mass = 150/g = 15.3kg
(120-150*uk)/15.3kg = 3m/s² => 120N - 150*uk = 3m/s²*15.3kg = 45.9N
(120N-45.9N)/150N = uk = 0.4938 <----------------
-
coefficient of kinetic friction is 0.5