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I saw this in a textbook and I want to confirm if it's true

If a body of mass m rolls down an inclined plane of height h and length l (hypotenuse), the work done is mgh. It is the vertical height of the inclined plane that is used in the calculation, not the length (hypotenuse) of the inclined plane.

Please how true is this because I believe we should be using a little bit of trigonometry since the object also moved a horizontal distance. Please if true, how?

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  • $\begingroup$ You can use trigonometry to relate the length $\ell$ of the hypotenuse to the height $h$ (e.g., $h=\ell \sin(\theta)$), but you don't have to because the height is already given. $\endgroup$
    – hft
    Commented Jul 8, 2023 at 23:04
  • $\begingroup$ The book excerpt is describing conservation of energy. If no dissipative forces are involved (e.g., friction), conservation of energy is path independent. $\endgroup$
    – David White
    Commented Jul 8, 2023 at 23:23
  • $\begingroup$ Thanks for your input. I now understand $\endgroup$
    – obanijesu adufe
    Commented Jul 9, 2023 at 14:08

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Assuming the incline plane is frictionless, and ignoring any air resistance, the only force doing work on the mass is gravity, and gravity acts downward. The amount of work done by gravity equals the loss of gravitational potential energy, $mgh$. Thus the work done equals $mgh$ and it equals the increase in translational and rotational kinetic energy of the mass as a result of the work done by gravity.

Hope this helps.

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  • $\begingroup$ Yes. Thank you very much $\endgroup$
    – obanijesu adufe
    Commented Jul 9, 2023 at 14:03
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With gravitational potential energy, all that matters is vertical height. Horizontal displacement is irrelevant. It's that simple.

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  • $\begingroup$ Thanks. I now understand $\endgroup$
    – obanijesu adufe
    Commented Jul 9, 2023 at 14:06
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Another way of thinking is to decompose the force of gravity in a component parallel to the incline and another perpendicular to it. The work of the net force is $F_g\parallel l$, because both the Normal and the component of the force of gravity perpendicular to the incline do no work due to being orthogonal to the displacement. (remember that work is the dot product of force by the displacement).

But $$F_g\parallel = mg\frac{h}{l} \implies w = mgh$$

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  • $\begingroup$ Thanks, I got more clarity with this $\endgroup$
    – obanijesu adufe
    Commented Jul 9, 2023 at 14:06

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