Question

an 88kg cardboard box is being pulled along a level floor. the coefficient of static friction between the box and the floor is 0.23, and the sliding friction is 0.12.
What is the weight of the box?
What force is needed to start the box moving?
What force is needed to keep the box moving at a constant velocity of 0.5m/s?
Once the box is moving what force must be applied to accelerate the box 2.5m/s/s?

Answer 1

a. Wb=m*g = 88kg * 9.8N/kg = 862.N.=Wt. of box.

Fb = 862.4 N. @ 0o. = Force of box.
Fp = 862.4*sin(00 = 0. = Force parallel to floor.
Fv = 862.4*cos(0) = 862.4 N. = Force perpendicular to floor.

Fs = u*Fv = 0.23*862.4 = 198.4 N.=Force
of static friction.
Fk = u*Fv = 0.12 * 862.4 = 103.5 N. =
Force of kinetic friction.

b. Fn = Fap-Fp-Fs = m*a.
Fap-0-198.4 = 88*0 = 0
Fap = 198.4 N. = Force applied to start
box moving.

c. Fn = Fap-Fp-Fk = m*a
Fap-0-103.5 = 88*0 = 0
Fap = 103.5 N. = Force applied to keep
box moving.

d. Fap-Fp-Fk = m*a.
Fap-0-103.5 = 88*2.5 = 220
Fap = 220 + 103.5 = 323.5 N.