*I really don't know how to get started. I have done lots of the homework, but this one (and a couple of others) are really giving me PROBLEMS.*

A stockroom worker pushes a box with a mass 10.5 kg on a horizontal surface with a constant speed of 3.00 m/s. The coeffecients of kinetic and static friction between the box and the surface are .330 and .400, respectively.

What horizontal force must the worker apply to maintain the motion of the box?
Answer expressed in N.

If the worker stops pushing, what will be the acceleration of the box?
Answer expressed in m/s^2.

I have difficulty with your opening statement, as this is some of the simplest force problems in existance. Perhaps you need to rexamine your efforts. Anyone can type a statement expressing what they think needs to be done.

Force friction= mass*g*coeffFriction

In the first case, to maintain the motion, the kinetic coefficent is used.
In the second case, the above force of friction is equal to mass*deaccaleration.

To solve this problem, we will use the formula for frictional force:

Force of friction = mass * acceleration due to gravity * coefficient of friction

First, let's calculate the force of friction when the box is moving with a constant speed. In this case, the coefficient of kinetic friction is used.

Force of friction = 10.5 kg * 9.8 m/s^2 * 0.330

To find the horizontal force the worker must apply to maintain the motion of the box, we need to overcome the force of friction. Since the box is moving at a constant speed, the applied force must be equal in magnitude but opposite in direction to the force of friction.

So, the horizontal force the worker must apply is:

Horizontal force = - Force of friction
= - (10.5 kg * 9.8 m/s^2 * 0.330)
= - 32.445 N (rounded to three decimal places)

Note: The negative sign indicates that the worker needs to apply a force in the opposite direction of the frictional force.

Now, let's consider what happens when the worker stops pushing. In this case, the box will experience static friction, which needs to be overcome for the box to start accelerating.

The force of friction when the box is at rest is:

Force of friction = 10.5 kg * 9.8 m/s^2 * 0.400

To calculate the acceleration of the box when the worker stops pushing, we can use Newton's second law of motion:

Force = mass * acceleration

The force acting on the box is the force of friction, and the mass is given as 10.5 kg. Therefore:

Force of friction = mass * acceleration
10.5 kg * 9.8 m/s^2 * 0.400 = 10.5 kg * acceleration

Simplifying:

9.8 m/s^2 * 0.400 = acceleration

acceleration = 3.92 m/s^2

Therefore, when the worker stops pushing, the box will experience an acceleration of 3.92 m/s^2.