A box of books weighing 290 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 34.9° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 3.80 m, starting from rest?

Answer: 1.95s

Explanation:

Given

ma = 290 cos 34.9 - fk

fk = 290 cos 34.9 - ma

fn = mg + 400 sin Φ

fn = 290 + 400 sin 34.9

fn = 290 + 228.9

fn = 518.9

fk = fn * uk

uk = 0.57

290 cos 34.9 - ma = 518.9 * 0.57

290 cos 34.9 - ma = 295.8

290 cos 34.9 - 295.8 = ma

ma = - 58

m = 290/10 = 29

a = 58/29

a = 2

Using equation of motion

S = ut +.5at²

3.8 = 0 +.5*2*t²

3.8 = t²

t = 1.95s