Question

Find the acceleration of a box of mass 3.0 kg, if applied force is 20.0 N at 37° above the x axis, and the coefficient of kinetic friction is 0.20.​

Answer 1

Hello:

First, the normal force in y axis:

∑Fy = 0

∑Fy = N - Fy

∑Fy = N - 20 N * sen 37°

∑Fy = N = 12,87 N

Formula:

R = ma

F - F' = ma

F - (μN) = ma

20 N - (0.2 * 12,87 N) = 3 kg * a

20 N - 2.574 N = 3 kg * a

17,426 N / 3 kg = a

a = 5,8 m/s²