A block of wood weighing 0.900 N rests on a table. A bullet weighing 0.100 N traveling with a velocity of 100 m/s hits the block of wood and sticks in it. The velocity of the bullet-block combination after the collision is 10 m/s. The coefficient of kinetic friction between the rough block and the table is 0.800. a) How far will the block slide before it skids to a stop?

Equation:

Ff = μ FN

1. The rocket would experience a downwards force due to gravity from the moon.  Also, the rocket would experience an upwards force from the acceleration of the rocket thrusters.  Assuming you are neglecting the force of air friction, there would be 2 forces.  If you are considering air friction as well, there would be also a downwards force of kinetic air friction acting down on the rocket as it flies, making it 3 forces.  If the rocket has already taken off, then there will be no normal force.  However, if the rocket is still on the surface of the moon, the rocket would experience a normal force from the moon.  Since the question isn't specific I can't provide a concrete answer, but it would range from 2-4 forces depending on the above factors.

2.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.

3.  Same idea as question 2.  First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N

So the person is exerting 144 N.  (Which I hope is a choice that you forgot to type out on the question)

5. Image 3 has the least amount of tension since both tension forces exerted on the cables are shared amongst the two cables and they are parallel to the downwards force of gravity that acts on the object and against them.