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Suppose a ball is moving in a horizontal projection with some velocity $u_0$ in the positive $x$ direction. Here we see that although gravity is perpendicular to the velocity vector, it changes the kinetic energy. Why is this so?

When I studied kinematics I had no problem with this type of problem because I would solve the $x$ and $y$ axes independently. But, in the chapter on work, power, and energy, I was told that a force perpendicular to the displacement vector would not do any work, i.e. change it's kinetic energy.

I may have kind of asked the same question I asked a couple of days ago, but please help me out here.

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  • $\begingroup$ I left a hint. If you are still confused after having done a more explicit calculation, I implore you to edit your post to include said calculation, and help can be given more appropriately from there. $\endgroup$
    – Relativisticcucumber
    Commented Sep 6, 2023 at 3:41
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Sep 6, 2023 at 3:54

3 Answers 3

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In the very first instant of its travel, when the ball's path is horizontal, gravity does no work on the ball. However, after this first instant the ball's motion is no longer horizontal, and gravity is no longer acting perpendicular to the ball's path. So if we integrate over any finite time interval, gravity does work on the ball during this interval, and so increases its kinetic energy by an amount $mgh$ where $h$ is the vertical distance travelled by the ball.

Note that this applies to ballistic motion. If the ball's motion is constrained so that it can only move horizontally (e.g. it is rolling on a smooth flat table top) then gravity does indeed do no work on the ball, and its kinetic energy is constant.

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  • $\begingroup$ I would suggest "motion not constrained to be exclusively horizontal" or something like that, rather than "ballistic". Lest pendula and simple machines hatch confusion in a month. $\endgroup$
    – g s
    Commented Sep 6, 2023 at 18:03
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Rephrase to "A force perpendicular to the displacement vector predicted after including that force in the equations of motion is not the means by which any work is done."

(Sum the forces, apply Newton's 2nd.)

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The work is done by the gravity not to the perpendicular displacement vector (displacemnt component along X axis) but to the parallel displacement vector (dsiplacement component along Y-axis).

Only point to understand here is that displacement vector is completely horizontal only at the instant the projectile is launched. Immediately after that it starts accelerating vertically downwards due to gravity and now has displacement vector with both horizontal and vertical components.

So what you studied in the chapter on work, power, and energy,(a force perpendicular to the displacement vector would not do any work) is still absolutely correct and so is your method of solving the kinematics problems by breaking into X and Y components.

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