Wb = 314 N.
m*g = 314
m = 314/g = 314/9.8=32 kg.=mass of box.
Fb = 314N[0o].
Fp = 314*sin(0) = 0 = Force parallel to floor.
Fv = 314*cos(0)+475*sin35 = 586 N. =
Force perpendicular to floor.
Fk = u*Fv = 0.53 * 586 = 310.6 N. = Force of kinetic friction.
a. Fap*cos35-Fp-Fk = m*a
475*cos35-0-310.6 = 32*a
389-310.6 = 32a
78.4 = 32a
a = 2.45m/s^2.
d = 0.5a*t^2 = 5 m.
0.5*2.45t^2 = 5
1.225t^2 = 5
t^2 = 4.082
t = 2.02 s.
A box of books weighing 314 N is shoved across the floor by a force of 475 N exerted downward at an angle of 35° below the horizontal.
(a) If μk between the box and the floor is 0.53, how long does it take to move the box 5.00 m starting from rest?
What is the maximum coefficient of kinetic friction between the box and the floor that allows the box to move from this applied force.
1 answer