A workerapplies a force of 138N on a heavy box by using a rope held at an angle of 23degree with the horizontal.

Question

A worker applies a force of 138N on a heavy box by using a rope held at an angle of 23 degree with the horizontal. If the mass of the box is 31kg, and coefficient of friction between the box and horizontal floor is 0.4, how much time the worker needs to move the box 15m.

Answer 1

F_G = mg = 31•9.81 = 304.11 N

The Force of the Pulling Man (F_P) = 138N can be broken into 2 parts F_Py and F_Px

F_Py = F_Psin(θ) = 138sin(23°) = 53.92 N

F_Px = F_Pcos(θ) = 138cos(23°) = 127.03 N

∑Fy = 0 = F_Py + F_N - F_G so F_N = F_G - F_Py = 304.11 - 53.92 = 250.19 N

F_F = μF_N = 0.4•250.19 = 100.08 N

∑F_x = ma

∑F_x = F_F - F_Px = ma

100.08 - 127.03 = 31a

-26.95 = 31a

a = -26.95/31 = -0.869 m/s²

Assuming the initial velocity is 0 and using the kinematic equation x = 1/2at² then:

-15 = 1/2(-0.869)t²

-15 = -0.4347t²

t² = 34.503

t = 5.9 s