Is work done dependent on velocity? If I lift an object off the ground to a particular height at a particular velocity is the work done in this action the same as if I lifted the object to the same height but at a greater/lower velocity? (Ignoring friction but taking into account gravitational pull on the object)
Thanks for any explanations.
 A: I wasn't going to post an answer to this, but the other answers seem very insufficient to me and may give the wrong idea.
The work required to lift an object off the ground is not affected by velocity directly.  The work is proportional to the product of the force acting on the object, and it's displacement in the direction of that force.
This means that if you are applying the same force over the same distance, it takes the same amount of work, regardless of the velocity.
The velocity is relevant to the power required; which is energy per unit time.  The faster you want to travel the displacement, the more power is required, so you need to deliver the same amount of energy in a decreased time frame.
You talk about overcoming gravity in the comments, and I will try to clear up confusion there as well.  If gravity increases, to overcome gravity you would need a greater minimum force to overcome it's effects.  Because of this, an increase in gravity will lead to an increase in required force for the same results, and if that force increases, the required work would increase as well.
So to reiterate again, velocity has no direct bearing on the applied work.  It will affect the required power, but the amount of work required to move the specified distance only depends on the displacement and the applied force; not upon how quickly it travels the displacement.
A: The work $W$ goes to potential and kinetic energy, $W=\Delta E_\mathrm{kin}+ \Delta E_\mathrm{pot}$. Of course if you lift your object from $z=0$ to $z=h$ very fast, and do not apply a negative work to fix the object at the height $z=h$, the work you apply also goest to kinetic energy at the final position $z=h$. But if you make sure your object has the same speed before and after your lift, then the work done in lifting an object of mass $m$ a height $h$ is $W$= $0+\Delta E_\mathrm{pot}=mgh$, no matter how the object got to that position $z=h$. But if you used a catapult and increased the objects speed by $\Delta v$, the work done in lifting the object a distance $h$ would certainly be larger by $\Delta E_\mathrm{kin}=\frac{1}{2} m (\Delta v)^2$.
A: If the final height is the same, the work done is independent of velocity. 
Suppose for simplicity a constant force during lifting. The minimum force is mg, and the work is mgh for that case.
Any bigger force results in an accelerated lifting, because there is a resultant force upwards: (F - mg > 0). In order to reach the same height, that force must be "turned off" before that point, and the weight will slow down until be at rest in the final height.
From the kinematic equation $v^2$ = 2ah:
h1 = 1/2 $v^2$/(F/m - g)  (accelerating 0 to v)
h2 = 1/2 $v^2$/g          (slowing down from v to 0)
Eliminating $v^2$ from the equations, and using that h2 = h - h1:
h1(F/m - g) = (h-h1)g
h1F/m = hg => h1 = hg/(F/m)
W = Fh1 = Fhg/(F/m) = mgh
So, the work is independent of velocity or force.
A: No the energy will be the sam because
W=mgh
But the power will be different
A: If the object is lifted to the height, $h$ but with having some initial velocity, $u$ and also thinking that inital and final velocities are same means the velocity of object when it reaches $h$ height remains $u$ again then, you are not doing any extra work on that object. The work done by you is $mgh$. Because the object was not accelerating, there is no extra force needed for lifting object. The force is same all the time and it is equal to $mg$.
A: Let’s say there are 2 boxes, same mass same volume same everything, needing to be brought to the same shelf. Box A will be lifted in 10 seconds while Box B will be lifted in 5 seconds. It may seem as though since the second one needs more force to push the box faster and more, the work done is unrelated.
Force = mass x acceleration
In Box A, let’s say the mass is 10kg, and there is no acceleration, it has a constant speed of 50cm/s maybe but there is no added acceleration, so it remains as 10.
For Box B the mass is the same, but the speed now is 1m/s, however, the acceleration is still the same, it moves at a constant speed so the force in both is still 10kg x 10m/s^2.
A: 
work done = force (f)× displacement

According to Newton's second law force is a product of mass and acceleration.

f = ma

Acceleration means rate of change of velocity.

Work done is directly proportional to that of velocity

A: Yes the work depends on the velocity as pointed out in other answer not directly but consider what happens when you throw a pebble vertically upwards. 
As the distance of the Pebble increases from the ground it's velocity naturally decreases (as a function of height) as you may know. 
But if you are insistent of having the pebble go upwards against the gravity with constant force, then you have to apply some king of external force for that to happen. So in this case work done is against gravity and by an external agent to maintain the constant velocity.
So yeah if you talk about direct formula work done is dependent on the distance travel and power is associated with power but you can find always  mathematical relation given the data. Also theoretically speaking it does depends on the velocity being constant in someway or other here. 
