Question

http://oi67.tinypic.com/29yig69.jpg
Any help?

Answer 1

I'm fairly certain the amount of force needed to move the box is going to be the maximum of the static frictional force, which again, is:
\[\huge F_s=\mu_s mg\]This should be the minimum force required to move the box

Once it's moving, there's a frictional force of\[\huge F_k=\mu_k mg\]We will need at least that amount of force to keep the box moving.

I would check my work with @ganeshie8 to be sure.

Answer 2

Looks good to me!
for second part, constant velocity means acceleration is 0, so we give the block just the necessary force to overcome the kinetic friction...

Answer 3

Fs= .502(2.03kg)(9.81m/s^2)
Fk= .273 (2.03kg)(9.81 ms/^2)

Answer 4

@Hoslos

Answer 5

@ganeshie8

Answer 6

@CShrix

Answer 7

Fs= 9.9969786
Fk=5.4366039