When the projectile is in motion what is the net work done on projectiles?

What i think? 0

first view: We know when we throw projectiles the initial KE and final KE is 0. So from work energy theoram the net work done must be 0.

second view: Lets see from the perpective from x direction and y direction. In X direction velocity is constant meaning there is no acceleration which in turn means force is 0 and in Y direction there is no displacement and $Work done = F \times displacement$ so it is zero in Y direction as well. So overrall net work done is zero.

Problem: when I googled i find many answer not written direclty 0. Here is a link http://answers.yahoo.com/question/index?qid=20101111093951AAtRsWP

I am currently reading GCSE mechanics and all the mathematical modelling is true such as neglecting air resistance . So lets see it from classical physics view only.


1 Answer 1


Yes, if you imagine a projectile shot from ground level, when it hits the ground again the total work done by gravity will be equal to zero.

To clear up a possible misconception: When you throw a projectile the initial and final kinetic energies are not zero (otherwise the particle wouldn't have any velocity and so wouldn't move!). However the initial and final kinetic energies are the same, and so the change in kinetic energy is zero.

If you look at the link you provided, that deals with a case where the projectile isn't shot from ground level - it's launched from above the ground. So when the projectile hits the ground the net y-displacement in this case isn't zero, and so the work done by gravity isn't zero after all.

If you want to look a bit more deeply into this topic, you'll find that the question of under what conditions forces do work or not over a path helps define the concept of a conservative force. It turns out that gravity is a conservative force, and so has various nice properties, such as not doing any work in this case.

  • $\begingroup$ If my answer helped, consider accepting it - it'll mark the question as resolved so people know you've sorted your problem. $\endgroup$ Commented Jul 17, 2015 at 10:23

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