I have learned that $$\mathbf{F} = m \mathbf{a}$$ where both the force $\mathbf{F}$ and acceleration $\mathbf{a}$ are vectors. This makes sense, since both force and acceleration have a direction.
On the other hand, the kinetic energy $$K = \frac12 mv^2$$ looks completely different. It doesn't seem to depend on the direction. How are these two concepts related?
Answer
When a force is applied over a certain distance, that force does mechanical work, $W$. If the force is constant $F$ and the object it is exerted on is moved by a distance $\Delta x$, then $W=F\Delta x$. If the force is not constant but a function of the position, this turns into an integral: $$W = \int_{x_1}^{x_2}F(x)\,\mathrm d x.$$ If you don't know calculus yet, just ignore this.
Note, that is is not important how long (in time) the force is exerted. E.g. a cup on a table will feel the constant force due to gravity but is won't move (because the table is pushing it upwards with an equal but opposite force) so there is no work done on that cup, meaning that its energy content won't change.
Work is basically just energy change. Depending on how the work is applied, it will increase (or decrease) a specific kind of energy. If the work leads to a change in the (absolute) velocity, it will modify the kinetic energy.
E.g. if a car accelerates from standstill with constant acceleration $a$ (i.e. the engine will exert a constant forward force on the car), its velocity increases linearly in time, $v(t)=vt$ and its position quadratically, $x(t)=\frac{1}{2}at^2$. After a time $t_1$, it went over a distance $x_1=x(t_1)=\frac{1}{2}at_1^2$, the work done by the engine will be $Fx_1 = max_1= \frac{1}{2}ma^2t_1^2$. At time $t_1$, the velocity of the car is $v_1=v(t_1)=at_1$, so we can write the work done by the engine as $\frac{1}{2}mv_1^2$. This is exactly the amount of kinetic energy gained by the car. So the work done by the engine was used to increase the car's kinetic energy.
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