Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
(1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:

The woman needs to pull the rope at 480.2N
Melting, of course. Just as how an ice cube melts to water.
1) 29.4 N
The force of gravity between two objects is given by:

where
G is the gravitational constant
M and m are the masses of the two objects
r is the separation between the centres of mass of the two objects
In this problem, we have
(mass of the Earth)
(mass of the box)
(Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)
Substituting into the equation, we find F:

2) 
Let's now calculate the ratio F/m. We have:
F = 29.4 N
m = 3.0 kg
Subsituting, we find

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol
.
We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

where a is the acceleration of the object. Re-arranging,

which is exactly equal to the quantity we have calculated above.
I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.
I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball <u><em>drop</em></u>, at the same time and from the same height as the first ball.
Now I'll go ahead and answer the question that I've just invented:
Strange as it may seem, <em>both</em> balls hit the ground at the <em>same time</em> ... the one that's thrown AND the one that's dropped. The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.
And here's another example of the exact same thing:
Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else <em>drops</em> another bullet at exactly the same time, from a point right next to the end of the rifle barrel. I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.