If you are asked to find the work done by gravity on a projectile, the force will obviously be the mass multiplied by gravitational acceleration. What will the displacement be, the horizontal or vertical one?


The work is given by:

$$ W = \int \vec{F}.d\vec{r} $$

where the force $\vec{F}$ and the displacement $d\vec{r}$ are both vectors and $.$ is the dot product. We can write $d\vec{r}$ as:

$$ d\vec{r} = d\vec{x} + d\vec{y} $$

where $d\vec{x}$ and $d\vec{y}$ are the displacements in the horizontal and vertical directions respectively, and substituting this into our expression for the work we get:

$$ W = \int \vec{F}.(d\vec{x} + d\vec{y}) = \int \vec{F}.d\vec{x} + \int \vec{F}.d\vec{y} $$

But the force $\vec{F}$ is directed downwards, i.e. in the $\vec{y}$ direction, so:

$$ \vec{F}.d\vec{x} = 0 $$


$$ \vec{F}.d\vec{y} = Fdy $$


$$ W = \int Fdy $$

and since $F$ is a constant we can take it out of the integral to get:

$$ W = F \int dy = F\Delta y$$

where $\Delta y$ is the $y$ distance moved. So the work is the force due to gravity times the vertical distance moved.

  • $\begingroup$ So in a projectile motion where the ball returns to starting height, the work done by gravity is 0? $\endgroup$ Jun 16 '14 at 11:26
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    $\begingroup$ @Gummybears: if work is done on the projectile its energy must change. Is the energy of the projectile at the moment it is launched different to its energy at the moment it returns to the ground (ignoring air resistance)? $\endgroup$ Jun 16 '14 at 11:30
  • $\begingroup$ Yes, sorry that slipped out of my mind. :D Thanks, and sorry for being so dumb. $\endgroup$ Jun 16 '14 at 11:42

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