# Why isn’t there a $\cos(\theta)$ in the work-energy theorem? [duplicate]

please don’t use any calculus in your answers I was reading the derivation of the work energy theorem in my textbook and it goes like this: $$W = Fs\Rightarrow W_{net} = mas$$ Since $$v^2 - u^2 = 2as \Rightarrow as = \frac{v^2 - u^2}{2}$$, substituting in the original equation, we get: $$W_{net} = \frac{mv^2}{2} - \frac{mu^2}{2}$$ Which is delta K.E.

But where did the $$\cos(\theta)$$ go which is there in the original equation of the work??

Shouldn’t the derivation be : $$W = Fs\cos(\theta)$$ And then we do the same as before and we should get: Net work $$= (1/2)m(v^2 - u^2)\cos(\theta)$$ , where theta is the angle between the NET force and the displacement, right? What is wrong with my reasoning? I know this probably isn’t correct because let’s take a simple case: $$v = \theta$$ and $$u$$ is a non-zero positive integer. Let’s say $$\theta = 180^\circ$$ i.e., according to my derivation, we would expect the work done to be negative, since $$\cos 180 = -1$$. But when you do $$v^2 - u^2$$, you would get a negative value itself which would then give us a positive net work which is incorrect. I think part of the wrong reasoning is that whenever we use the work equation, we only plug in the MAGNITUDES of the force and displacement as the positive or negative work is only decided by the term cos(theta). But then how do I get $$v^2 - u^2$$ as negative? If $$u$$ is let’s say $$10$$ m/s then $$v^2 - u^2$$ would be $$-100$$. I can’t just write +100 for that. Please clear my confusion.

• The scalar equation $W=Fs$ assumes that the force $F$ and the displacement $s$ are in the same direction (or, if you prefer, it assumes a one-dimensional scenario, so there is only one dimension). So implicitly it assumes that $\theta$ is $0$ and so $\cos \theta = 1$. Commented Apr 29, 2020 at 11:04
• Does this answer your question? Why is work scalar and the dot product of force and displacement?
– user258881
Commented Apr 29, 2020 at 11:08

If you apply a force on a body at θ to its displacement, magnitude of velocity(speed) only changes when you apply a acceleration parallel to its displacement vector ( v = ds/dt) so, here acceleration according to your equation is a cosθ. When you put this acceleration to your equation and multiply both side by mass gives you the same result as work energy theorem. Force(acceleration) perpendicular to displacement only changes particle's direction. So net work done by perpendicular force gives same result that is 0 ( w = FS cos90)

It will be hard not to use any calculus to derive the equation, by I will try my best to give you reasoning why is it so.

Work is defined as force component in line with displacement direction. That's where the $$cos\theta$$ term is coming from. It may be easier to see from this picture why is the case:

Our direction of displacement denoted by $$\Delta\vec s$$ is given by gray arrow. To take component of force in direction of that displacement, means taking length of dashed green line. Simple trigonometry tells us, that this value is $$F \cos\theta$$

Let's agree to denote $$F\cos\theta$$ as $$F_{actual}$$ I.e. our new force is simply old one corrected by $$\cos\theta$$, which from above assumptions tells us, that this is a force parallel to displacement.

Since F=ma, $$\Delta s=\frac{v^2-u^2}{2a}$$, we have:

$$W = F_{actual}\cdot \Delta s=ma\cdot\Delta s=m\biggl(\frac{v^2-u^2}{2}\biggr)=\Delta KE$$

In short, our $$\cos\theta$$ gets 'sucked' in F term.

This is however a very simplified model, and I would strongly encourage you to try get familiar with more rigorous, actual equation. It is not as bad as it seems!

• What if cos(theta) gives me a negative value so Fcos(theta) becomes negative but I’ve chosen the direction in which the force acts as positive? Let’s say Fcos(theta) = -10 . If in my sign convention, I choose the direction in which the force acts as positive, do I ignore the negative and just write Fcos(theta) = ma = 10? Commented Apr 29, 2020 at 12:53
• Indeed force component can be negative. Simple scenario is you simply move against some force. As far as work equation is concerned, you need to add components of force vector parallel to displacement. It does not matter if it is positive or negative, as long as it is 'in line' with displacement. Can you deduce, what sign does work has to be from work energy theorem if particle starts with some velocity u and ends up with zero velocity? What does it tell you about net force direction during particle 'journey'? Commented Apr 29, 2020 at 13:02
• Another question: why is it that we only plug in the MAGNITUDES of the force and displacement in the work = fscos(theta) equation? Commented Apr 29, 2020 at 13:12
• Work is a scalar quantity. Force and displacement are a vector quantities. In general work is defined as $dW=\vec F\cdot d\vec s$, which translates to ‖F‖ ‖ds‖ $\cos\theta$ where the operator is called a dot product. Commented Apr 29, 2020 at 13:33

I hope you know about the components of forces... if not read this part : Any vector quantity can be separated as sum of two different vectors. That means a force applied along North East direction can be seen as two forces, of same strength one along East and the other along North acting together. (If one of them was stronger, the net force would not be along NE, but tilted towards the stronger side).

Answer : When we say $$W = F.s.cos\theta$$ , $$Fcos\theta$$ is the component of Force along direction of s.

Your derivation dealt with a particle moving along a straight line under a constant acceleration. Let us take the same example... Consider a particle that can move only along $$X$$ axis. You apply a force in some random direction and you find that the particle is moving. After some time, you see that the particle has displaced by $$s$$ distance.

Understand that acceleration along $$s$$ direction is only due to force along s direction which is $$Fcos(\theta)$$. $$F$$ as a whole does not cause acceleration along $$s$$ direction. Thus, when applying newton's law, it is $$Fcos\theta = ma_s$$.

• Thank you!!!!!!!! Yours is probably the answer I was looking for. I did not realise that Fcos(theta) = ma and just substituted F=ma Commented Apr 29, 2020 at 13:22
• There is a reason PSE has MathJax ;)
– user87745
Commented May 6, 2020 at 16:44