1)
Draw a simple diagram showing the box and the forces that act.
The weight of the box (500 N) is also the size of the reaction force, given the name "Normal force" because it acts perpendicular to (ie normal to) the platform. This Normal force supports the box weight.
The value of the static friction force (which acts horizontally) is given by:
Static Friction Force (SFF) = (us) x Normal force
where us = coefficient of static friction
SFF = 500(us)
Normal force = 500 N (vertical)
Motion force = 200 N (horizontal)
In order to push or pull the box into motion, the SFF which is a reaction force must be overcome by the motion force. At the value of 200 N the motion force is just able to overcome SFF.
So SFF = 200 N when the size of the motion force starts box moving.
and
we already know that
SFF = 500(us)
setting them equal
500(us) = 200
us = 200/500 = 0.4 ANS (a)
when the box is moving the force of friction becomes a value given by the kinetic coefficient of friction = uke
Kinetic Friction Force (KKF) = (uke)(Normal force)
KFF = (uke)(500)
and
We are told a motion force = 160 N maintains constant speed motion of the box. In order to have constant speed motion there can be no acceleration (in the direction of motion). In order to have no acceleration there can be no NET force in the direction of motion. We know that there is a motion force = 160 N so the KKF which opposes this force must be equal in size to it,
KFF = 160 N
and
KFF = uke(500)
setting equal
uke(500) = 160
uke = 160/500 = 0.32 ANS (b)
2)
Work = Force x Distance in same direction as force
Work = 250 x 20 = 5000 J ANS (a)
Work = Force x Distance in same direction as force
Force and Distance are not in same direction
Force Cos 30 = component of Force parallel to ground
250(0.866) = 216.5 N = Force component parallel to ground
Work = 216.5 x 20 = 4330 J ANS (b)