To lift an object, do we need a force equal to its weight, or greater than its weight? We have all heard people saying that to lift an object of mass $m$, you have to apply a force $F$ equal to its weight $mg$. But isn't it getting the force equal to its weight from the surface to which it is attached to (normal force). Why it is willing to change that equilibrium state by getting the same force from us as from the surface? (Consider the situation devoid of any resistance) . I think we must be applying slightly more force to it in order to move it even with constant velocity at least at the start and balancing the force of gravity afterwards.
 A: In real situations, when a mass is sitting on a table, its weight will slightly flex the table. So lifting with a force that equals the weight will remove normal force pressure from the surface, the surface will flex back to it's former position giving the slight increase in net upward force to start acceleration. This can be visualized easier if you imagine the mass resting on a spring. So in most real situations applying a force equal to it's weight would lift it.
A: You are correct that there must initially be a net upward force, no matter how small and how brief, to get the object going. But in addition to balancing the forces immediately after to achieve constant velocity there must be a net downward force just before reaching the height such that the object will come to rest, if it is to possess ONLY gravitational potential energy at that height. Otherwise the object will possess both gravitational potential energy and kinetic energy when it reaches that height.
Hope this helps 
A: There are two points to be clarified here. 


*

*The normal reaction force from the surface is a self-adjusting force. In particular, it can take any value so as to prevent the object in contact from penetrating. So, if an object resting on a surface has a weight $w$ then the normal reaction force would be $w$ in the upward direction. Now, if you apply an external upward force on the object (with your hand, say) of a magnitude $w/2$ then the normal reaction force from the surface would change its value to $w/2$. Now, if you apply an external force of a magnitude $w$ in the upward direction then the normal reaction force from the surface would reduce to zero.

*However, as you correctly notice, when the upward external force is exactly the same as the weight in magnitude, the object is still in perfect equilibrium. And since the initial velocity of it was zero, its velocity would still remain zero because equilibrium means no acceleration. So, there would be no movement. So, in order to actually lift the object, you do need to provide an upward force which is at least slightly greater than the weight of the object. Once you apply such a force even for a tiny amount of time, the object would pick up an upward velocity because it would have been subjected to an upward acceleration for that tiny amount of time. Once this is accomplished, you can reduce the magnitude of the upward force to be exactly the same as the magnitude of weight and the object will continue to move in the upward direction, in equilibrium, but now, with a constant velocity (that it picked up during that tiny amount of time of acceleration). 

A: All of the above answers make a false assumption. They assume that the object is not currently moving vertically. This is not given in the question. So the correct answer should be it depends on whether the object is moving vertically or not.
If the object had already been propelled upwards and we "caught" it midflight the "very small force for a very small time in order to overcome equilibrium argument" doesn't hold.
A: Your conclusion is correct. To make the object move you must initially apply a force greater than the weight of the object in order to accelerate it from rest. After that you can reduce the force to match the object's weight and it will continue to rise at whatever speed it had reached while you were accelerating it. 
In practice it will be impossible to apply a force that is at all times exactly equal to the weight of the object, so what will actually happen is that the object's speed will vary and you will continually have to make small corrections to the force.
If you want a more precise answer, you should also take into account:
1) Air-resistance, the effect of which which will depend on the size and shape of the object, its speed relative to the air through which it is moving, and the pressure and moisture content of the air, etc.
2) Buoyancy, the effect of which which will depend on the volume of the object and its density.
A: Assuming the object is on the surface of the earth there is a slight gravity gradient, so applying a constant force equal to the weight of the object at that initial position results in an unstable situation. 
A slight movement of the object upward (from thermal motion, vibration or whatever) will reduce the force of gravity by about 0.3 mGal/m -- a gradient of 3,080 E (eotvos), so absent any other significant forces (a rather big assumption) it would tend to eventually fly off into space. 
A: It really depends on your definition of "lift".
If lift is taken to mean that the object should cease applying any pressure to the surface underneath it, then the force that you need to apply needs to be exactly equal to its weight, no more.
If lift is taken to mean that the object should not only cease applying any pressure to the surface underneath it, but also move away from the surface, then of course you will need to apply such a force that not only overcomes the weight of the object, but also accelerates it by a tiny little bit so that it may start to move away.  Without at least a tiny bit of initial acceleration the object will not move.
(In the actual, real™, physical world the object will of course move away from the surface even if you try the first scenario, partially because of air currents, vibrations, etc. and partly because you cannot actually apply a force which is exactly equal to a weight: it will necessarily vary slightly over time due to vibrations in your lifting apparatus. So, while the force that you are applying is less than the weight of the object, you will see no effect, but at the first moment that the force ever so slightly exceeds the weight of the object, the object will move.)
A: On the surface of a planet with an atmosphere, you have two things helping you: The buoyancy of the object (basically the weight of the air that it replaces), and assuming the planet is rotating and you are not at one of the poles, the centrifugal power. 
So to lift an object, you only need to apply a force slightly less than the weight, because there are two effects that help you. 
This will be very noticable if you try to lift a balloon filled with air. The air weighs a few grams, but you don't need to supply a force for that. With 100 kg of iron, only about 12 liters of air weighing about 15-16 grams are replaced, so the difference is tiny. 
PS. See Adrian's answer for a third force helping you.
PS. Someone could claim that all these forces should just be added to the force that you apply yourself. Centrifugal power is different. It's not a force, it's just an effect of the object trying to move in a straight line, when due to earth rotation for example all objects resting on the surface actually move at very high speed along a circle, not a straight line. 
A: From a computational point of view, a force is a floating point number, not an integer. With floating point numbers, there is not much point in saying things like x equals y, because x and y may differ a very small amount, so small that some consider this as equal, and others may consider this as different. If you program:
float x;
....
if (x == 1.23) ...

then the if-clause may never get true, because x may get very close to 1.23, but never equal to it. Normally you would instead code something like:
if (x <= 1.23)

or:
if (x > 1.22 and x < 1.24) ...

For your question, if you want to 'lift' an object with mass m with a very small accelleration a, the required upward force F = m x a may be very small, where some may consider this as "zero" or "no" force and others may consider this as a considerably non-zero force.
It all boils down to what you mean with "equal".
A: Maybe you've heard of Isaac Newton's second law of motion...
$F=ma$
where
$F=$ net force on an object
$m=$ mass of object
$a=$ acceleration of object
In your question, you ask about a situation where the net force on the object is zero.  The force due to gravity (weight) is equal and opposite to your lifting force.  You can see using the equation that if $F=0$, then also $a=0$.
The object cannot begin to move with an acceleration of zero.  Therefore it will remain stationary.  If your lifting force is greater than the weight, then the net force will be greater than zero and the object will begin accelerating upward.  The magnitude of the acceleration will depend on both the net force and the mass of the object.
A: Yes we apply a slight greater amount of force at the beginning and then we lift it up in equilibrium condition.In fact in extremely small amount of time the object gains some velocity when we apply a force just greater than its weight.After that extremely small period of time we can lift the particle up in equilibrium.
A: Simplistically but hopefully usefully:
Apply Newtonian principles strictly.
Assume ideal conditions but include as many ideal factors as needed to be 'happy' with the result. eg 


*

*include or exclude buoyancy as desired - it is a real world factor but typically amounts for around 0.1% of the total "weight". eg for water air-buoyancy is about 1.2 grams per kg of water at sea level.  

*include or exclude drag losses when moving as desired - drag in air is about 0.6 x A x V^2 in Si units. (A - projected frontal area, V - velocity). At say 1 m/s that's around 0.5 newton per square metre of frontal area - not something you will easily notice in many scenarios. At 10 m/s it's around 50 newton and 'you may begin to notice'. t 100 m/s it's around 5 kN and will dominate in many real world scenarios. 
A "lifting" force exactly matching a supporting force replaces it seamlessly. 
Newton says so.
Believe him.
He knew what he was about :-)  
Any 'upwards' forces greater than the 'downwards' weight force will accelerate the body using standard Newtonian expressions. 
However - If we deal separately with the above weight versus "supporting force" balance, even though it is 'correct enough', it actually complicates what seems to happen. So once we are happy with the separate "hovering" and acceleration forces metaphors, you can then sum together all the components and deal with the 'vector sum'. So sum weight (= m.g), buoyancy (if considered, = volume a air-density), air-drag (gets messy quickly - best left ideal at zero to start with ~= 0.6 x Area x velocity squared) and the results will be exactly as you'd expect. (That's assuming you don't want to add in solar radiation pressure or any other 3rd 4th ... order effects).
QED?
A: As has been mentioned in @Adrian's answer in the real world there is nothing rigid, neither the support where the mass m is resting on be it a table or cement bench nor the mass itself.
So there is always a certain amount of energy reserved in the support due to its deflection/settlement under the weight of the mass.
Therefore there are two forces here, let's call them,
$$ F_1 = \text {the force we exert} \\ and, \quad F_2= Kx  $$
With K as spring constant of the support and x as the settlement of the support under the m*g.
So we actually need to apply just enough force (slightly less than the weight, mg) to give the spring action of the support a chance to push the object up act as a projectile moving up. And we can calculate the F1 just short of weight enough so that$$ W*H= F_1*H +\frac{1}{2}Kx^2 \\ F1= W-\frac{1}{2H}Kx^2  $$
With W as weight and H as height. And we inspect that the F1 is smaller than the wight.
And therefore we do not need any downward transient force at the top end of the trajectory.
A: The same thing was bothering me for a while. I knew that as we accelerate the object, it's gonna have to be decelerated as we end our lift. So the work done is simply the height we lift it times the average force. But not being able to prove it wasn't at all satisfying to me and it bothered me for a while. But recently, I think I managed to figure it out. Before I begin, though, I want to say that I know this post has been out for a while, but I haven't seen anyone show this kind of "proof" before, so I wanted to share it. If there's something wrong with it, let me know. And secondly, I am gonna be using calculus for this so make sure you're ok with that.
Ok, let's begin by saying we have a net force that is acting on the mass. This net force is the combination of the force that is lifting the object and the weight force of the object
$F_{net}(s)=F(s)-mg$
We're subtracting by the weight force because it's going in the "down" direction, or we can say the "negative" direction. Now let's use the very definition of work and see where it gets us with this net force (integrating from 0 to s). But, we're gonna be integrating the net force, not the force that is applied to the mass. So we'll use a different variable, $W_0$ instead of $W$
$W_0 = \int_0^s F_{net}(s) ds$
We're not integrating from 0 to h, but rather from 0 to s, you'll see why later. Now, we know that:
$F=ma$ and $a=\frac{dv}{dt}$, so:
$W_0 = \int_0^s m \frac{dv}{dt} ds = m \int_0^s \frac{ds}{dt} dv$
$= m \int_0^s v dv$
And since we're lifting it by a distance of s, you can imagine after the force stops lifting, the mass will have a velocity we'll call $v_0$, this is because the net force can have any function. So the net force can for example still be accelerating the object and then abruptly stop. If you start lifting something and then stop without any slow enough deceleration, the mass will keep going up in the air even after you had already stopped lifting it. So:
$W_0 = m \int_0^{v_0} v dv$
Now if we finally integrate this and plug in the bounds, we get:
$W_0 = \frac{1}{2} m v_0^2$
You might for sure recognize this, it's the kinetic energy of the mass, we're just calling it $W_0$. Now, since we know the velocity's gonna be positive cause the mass is going upwards, if we solve for the velocity we get:
$v_0 = \sqrt{\frac{2W_0}{m}}$
Now we're gonna be using our kinematic equations where $h$ will be the total height of the object, including the height it was already raised by, $s$:
$h = -\frac{1}{2} gt^2 + v_0t + s$
$v = -gt + v_0$
we're trying to solve for the highest height the mass reaches, so $v$ will be zero. And since we know what $v_0$ is:
$0 = -gt + \sqrt{\frac{2W_0}{m}}$
Solving for $t$, we get:
$t = \sqrt{\frac{2W_0}{mg^2}}$
Plugging this into our upper kinematic equation, we get:
$h= - \frac{1}{2}g \sqrt{\frac{2W_0}{mg^2}}^2 + \sqrt{\frac{2W_0}{m}} \sqrt{\frac{2W_0}{mg^2}} + s$
Simplifying this by canceling a bunch of things, we get:
$h = \frac{W_0}{mg} + s$
solving for $W_0$, we get:
$W_0 = mgh - mgs$
Now let's solve our first integral differently, by using:
$F_{net}(s) = F(s) - mg$
we get:
$W_0=\int_0^s \left(F(s) - mg \right) ds = \int_0^s F(s) ds - \int_0^s mg ds$
Now our second integral is easy since everything in it are constants, we can easily integrate it, plug in our bounds and get $mgs$, but our first integral seems a little more complicated. But don't worry, if you think about it, the first integral is simply the work done by the actual force that's lifting the object. This is what we're after, and we can simply, therefore, call it $W$. So what we get is:
$W_0 = W - mgs$
But recall from when we solved for $W_0$ last time, we got:
$W_0 = mgh - mgs$
Since both of these are equal, we can say that:
$W - mgs = mgh - mgs$
If we cancel out the $mgs$'s, we finally get that:
$\therefore W = mgh$
Where $h$ is the highest point the mass was at. So for example, applying a constant force that's bigger than the weight force of the mass is gonna have the mass end up at a higher height than at which you stopped lifting it. But if you add the height you raised it by with the height it went above the height you raised it by and then multiply the result by the weight of the object, you'll get the same amount of work done as you would if you were to gently lift the mass up by first accelerating it and then decelerating it.
