Tuesday

July 29, 2014

July 29, 2014

Posted by **John Miller** on Monday, April 8, 2013 at 9:23pm.

(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.

- physics -
**Henry**, Tuesday, April 9, 2013 at 10:16pmWb = 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.

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