We got a long way with this equation, which came from one of the constant acceleration equations in the y-direction.

½m v_y² = ½m v_oy² + ma_y Dy

We can do the same thing in the x-direction, to get:

½m v_x² = ½m v_ox² + ma_x Dx

Are we adding vectors or scalars here?

Another thing we can do is to add the equations. This gives:

½m v_x² + ½m v_y² = ½m v_ox² + ½m v_oy² + ma_x Dx + ma_y Dy

Let's simplify this a bit:

½m v² = ½m v_o² + ma_x Dx + ma_y Dy

Now, ma_x is the x-component of the net force, and ma_y is the y-component of the net force, so:

½m v² = ½m v_o² + F_NETx Dx + F_NETy Dy

Is there some way to simplify this? We also need to keep in mind when a term like F_NETx Dx is positive and when it is negative.

The equation can actually be written as:

½m v² = ½m v_o² + F_NET · Dr

What should we call this new quantity F_NET · Dr? It is some kind of energy.

Let's call it the work done by the net force.

This work done is:

• positive if the net force has a component in the direction of the displacement
• zero if the net force is perpendicular to the displacement
• negative if the net force has a component opposite to the displacement