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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.

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    $\begingroup$ The work (energy required) is the same. The power (energy per time) is proportional to velocity. $\endgroup$
    – Marius Ladegård Meyer
    Commented Dec 10, 2019 at 15:31
  • $\begingroup$ Thanks for your comment. So even if you apply a greater force to accelerate faster against gravity the work remains the same? $\endgroup$
    – dillib
    Commented Dec 10, 2019 at 15:41
  • $\begingroup$ Actually when you lift it at a some velocity you are actually providing two kinds of energy two it: Gravitational potential energy and Kinetic energy also. So higher the speed more the K.E. and hence more the work. $\endgroup$
    – user240696
    Commented Dec 10, 2019 at 15:49
  • $\begingroup$ Whoa, that's not the same situation. If you use a greater force, or the distance of displacenent increases, or both, then the work changes. But it is the force that matters, not the velocity. $\endgroup$
    – Marius Ladegård Meyer
    Commented Dec 10, 2019 at 15:53
  • $\begingroup$ @MariusLadegårdMeyer Do you not require a greater force to overcome gravity in order to accelerate to a greater velocity though? I have no idea what I am talking about so excuse my ignorance. $\endgroup$
    – dillib
    Commented Dec 10, 2019 at 15:56

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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.

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  • $\begingroup$ but since the OP mentioned lifting an object off the ground (presumably from rest) then wouldnt a greater velocity mean a proportionally higher work done?$$m\frac{\Delta V}{t} X distance=work$$ given that time,mass and distance are constant of course. $\endgroup$
    – Hisham
    Commented Dec 10, 2019 at 18:35
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    $\begingroup$ @UbaidHassan It doesn't say starting from rest, and I didn't see that as the question OP was asking, especially based on the title. I feel I covered what you talked about quite well anyways, because I was very explicit that I was referring to applying the same force, whereas this is the case where the force would change based on the velocity. OP's wording made me think he was talking more about a constant velocity situation, and I included enough about the forces that if that is not the case, this still shouldn't mislead them. $\endgroup$
    – JMac
    Commented Dec 10, 2019 at 18:48
  • $\begingroup$ So just to clear something up, if we had 2 objects, they both exerted the same amount of work, but with differing accelerations. Would gravity eventually bring them to rest at the same point and at the same time? $\endgroup$
    – PsyFi
    Commented Dec 11, 2019 at 0:23
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    $\begingroup$ @Dandan That seems like a really broad statement that I wouldn't say is true. I'm not really sure what in this answer makes you think that though, so it might be good to clear up. $\endgroup$
    – JMac
    Commented Dec 11, 2019 at 0:47
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    $\begingroup$ @Dandan As long as the product of force and displacement in the direction of that force is the same, you've applied the same work. The same force might accelerate one mass quicker than another through some distance, while applying the same force, because the less massive object accelerates more with the same force; but the change in potential and kinetic energy would be the same as if you applied the same force to a more massive object for the same displacement; even if it takes longer and the more massive object ends with less velocity. $\endgroup$
    – JMac
    Commented Dec 11, 2019 at 1:20
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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 $\Delta E_\mathrm{kin}=\frac{1}{2} m (\Delta v)^2$.

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  • $\begingroup$ Great explanation. Thanks for additionally explaining what happens with regard to kinetic energy if the starting and ending speed are the same. $\endgroup$
    – dillib
    Commented Dec 10, 2019 at 21:18
  • $\begingroup$ @dillib I think it's worth noting that if you look at the force instead of velocity, the net work is still $W = F \cdot \Delta x$ regardless of the velocity at the start and the finish. The point of that work formula is that the force times displacement actually ends up being the same as the change in potential and kinetic energy in a conservative field. $\endgroup$
    – JMac
    Commented Dec 11, 2019 at 0:56
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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.

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  • $\begingroup$ So when you say work is independent of velocity or force here, what are the factors that it would depend on? Particularly in this scenario you described? $\endgroup$
    – PsyFi
    Commented Dec 11, 2019 at 0:17
  • $\begingroup$ @PsyFi In this scenario (lifting a weight to a given height) it depends only on mass and height (assuming that variations of g are negligible). $\endgroup$
    – Claudio Saspinski
    Commented Dec 11, 2019 at 17:40
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No the energy will be the sam because W=mgh But the power will be different

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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 $mg$.

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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.

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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

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  • $\begingroup$ thats only if the mass of the object, the time taken in the acceleration of the object and the distance travelled are constants, ide like to add. $\endgroup$
    – Hisham
    Commented Jan 9, 2020 at 23:55
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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.

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  • $\begingroup$ It is a special case in which you are considering the final velocity to be 0. But in actuall the work done is dependent on $v^2-u^2$ not just u. $\endgroup$
    – user316791
    Commented Jan 23, 2022 at 15:57

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