posted by Uzy .
A 1,440-N crate is being pushed across a level floor at a constant speed by a force vector F of 380 N at an angle of 20.0° below the horizontal, as shown in the figure (a) below.
(a) What is the coefficient of kinetic friction between the crate and the floor?
μk = Correct: Your answer is correct.
(b) If the 380-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
(a) You claim you already have the correct answer for ìk.
(b) The friction force is
Ff = (ìk)*(M g -380 sin20)
Net horizontal pulling force
= 380 cos20 -Ff
Solve for a. M = 1440/9.8 = 146.9 kg