0
$\begingroup$

Why is

Work $=$ Newtons $\times \cos (\theta) \times$ Distance

and not

Work $=$ Newtons $\times \cos(\theta)$

My understanding is that Work is Newtons directed horizontally so why not just project the Newtons of the original force in the horizontal direction. Why multiply by the magnitude of the distance vector? (Assuming we don't already know it's a Dot product).

$\endgroup$
2
  • 2
    $\begingroup$ What are the units of $\text{Newtons}\cdot\cos(\theta)\cdot\text{distance}$? What are the units of $\text{Newtons}\cdot\cos(\theta)$? What should the units of work be? This doesn't answer the question, but it can certainly tell you when something has gone wrong... $\endgroup$
    – Xander Henderson ♦
    Commented Sep 18, 2017 at 17:44
  • $\begingroup$ By definition, $$W = \int \mathbf f \cdot d\mathbf l$$, Take $f$ constant and evaluate the integral. $\endgroup$
    – user312097
    Commented Sep 18, 2017 at 17:48

1 Answer 1

Reset to default
5
$\begingroup$

That's just what the work formula is: $\text{work = force $\times$ distance}$.

The force multiplied by $\cos\theta$ just gives you the horizontally directed piece of the force. It doesn't tell you how far that piece is directed, i.e., it doesn't tell you the distance. You need to multiply by the magnitude of the vector to get this distance.

My understanding is that Work is Newtons directed horizontally so why not just project the Newtons of the original force in the horizontal direction.

Work is not necessarily in the horizontal direction. Work can be done in any direction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .