What is the approximate weight of a 5.0-kilogram object at the surface of the Earth?

Weight is given by #W=mg#, where #m# is the mass of the object and #g# is the free-fall acceleration, #9.8m/s^2#. Note that #W# is also the symbol used to represent work. I'll give a short answer followed by a more in-depth explanation if you're interested to know where all of this comes from.

Given #m=5.0kg# and #g=9.8m/s^2#, we can calculate the weight of the object:

#W=(5.0kg)*(9.8m/s^2)#

#W=49N#

The weight of the object is #49N#.

On the surface of Earth, (assuming a level, horizontal surface), the the weight of an object is equal to the normal force that the surface exerts on it in order to support its weight.

Explanation 2.0:

A basic force diagram for an object at rest on a flat, horizontal surface:

Where #vecn# is the normal force and #vecF_g# is the force of gravity.

The net force on the object is zero; it is at rest and therefore has zero acceleration.

If we constructed a sum of forces statement, we would find:

#sumvecF_(n e t)=sumvecF_y=vecn-vecF_g=m*veca#

Note that there are only vertical forces acting on the object, none horizontally, so the net force is equal to the sum of the vertical forces.

Since #veca=0#:

#vecn-vecF_g=0#

#=>vecn=vecF_g#

Because #vecF_g=mg# for an object at rest,

#vecn=mg#

This is where the equation for weight of an object at rest comes from. When you stand on a scale, your weight is given by the normal force that the scale exerts upwards on you to support your weight, caused by the force of gravity pushing downward on you and the scale.

Note, however, that for an object which is not at rest, we can no longer use this simple equation. When we introduce acceleration, things get a bit more complicated. A typical example places a person standing on a scale in an elevator. If the elevator cable breaks and the person and elevator begin to fall downward, what is the reading on the scale?

We would have:

#sumF_y=vecn-vecF_g=mveca#

Now the acceleration isn't zero, but is equal to #-g#, since the elevator is in free-fall.

#vecn-vecF_g=-mg#

The force of gravity is still given by #F_g=mg#

#vecn-mg=-mg#

What happens when we try to solve for #vecn#, the normal force which represents the person's weight?

#vecn=0#

Both the person and the elevator accelerate downward at the same rate. This is why you would feel weightless if you experienced free fall. The feeling of weight you have on the ground is due to the support (normal) force of that surface. This is also why you would feel heavier if the elevator suddenly accelerated upwards.