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enter image description here Work is supposed to be force x distance, how did we end up with work equaling the difference of hights x gravitational acceleration ? Sorry for my poor understanding :D

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  • $\begingroup$ Gravitational acceleration Acts in y direction. $\endgroup$ Mar 21 '16 at 18:17
  • $\begingroup$ yeah I agree... But shouldnt it be a dot product between d and F ? $\endgroup$
    – therue
    Mar 21 '16 at 18:18
  • $\begingroup$ Work done is not force x displacement .. Work done is the dot product of force and displacement . If you are a beginner i will say work done is equal to the force x (displacement in direction of force) so as gravitational force acts in y direction we have to see displacement in y direction which is difference in heights and then you will get work done :) $\endgroup$ Mar 21 '16 at 18:22
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    $\begingroup$ Aah if think I got it ... The work is a dot product of force ans displacement, the gravitational acceleration has only Y components, the Y component of Distance is the cosine and the cosine between them is equal to the difference in hights ?? $\endgroup$
    – therue
    Mar 21 '16 at 18:28
  • $\begingroup$ Yes you are correct :-) $\endgroup$ Mar 21 '16 at 18:30
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Look at the forces in your system:

  • In the y-direction, you have gravitation force.
  • In the x-direction, there is no force.

Edit:

Suppose you have vectors for the force and the displacement: $F=(F_x,F_y)$ and $d=(d_x,d_y)$, now to get the work W, you calculate the dot product* of these two:

$W=F*d=F_x*d_x + F_y*d_y$

With

$F_x=0$ (because there is no force on your ideal frictionless inclined plane) and

$F_y= -m*g$ (gravitational force; m=mass, g=gravitation constant=$9.87 m/s^2$)

above equation will give you:

$W=F_y*d_y = -m*g*d_y$.

So we see, since no force acts in the x-direction, no work is done in the x-direction.

*=in genereal this would be a line integral, but it boils down to the same idea.

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  • $\begingroup$ im not sure I understand... can you draw a sketch ? $\endgroup$
    – therue
    Mar 21 '16 at 18:16
  • $\begingroup$ This can be a good comment! But not satisfying the quality of answer $\endgroup$ Mar 21 '16 at 18:16

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