A stock room worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 4.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.25. a) What horizontal force must be applied by the worker to maintain the motion? b) If the force calculated in parts (a) is removed, how far does the box slide before coming to rest

Answer 1

a = 0
so the sum of forces is 0
F is applied
- mu m g is friction force
so
F = mu m g
= .25 * 11.2 * 9.81

then
m a = mu m g
a = mu g
solve for a

time to stop first
0 = 4.50 - a t
solve for t

then

x = Xo + Vi t + (1/2) a t^2
x = 0 + 4.5 t - .5 a t^2

Answer 2

Vfinal = vi + at not negative it should be + sign i don't know pla reply