Posted by **Yosh** on Wednesday, December 8, 2010 at 7:18am.

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

- physics -
**Damon**, Wednesday, December 8, 2010 at 8:46am
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

- physics -
**Why negative**, Monday, August 26, 2013 at 4:09am
Vfinal = vi + at not negative it should be + sign i dont know pla reply

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