Newton's laws vs energy for solving a problem I have a problem I solved using kinematics/Newton's 2nd law. 

It gives the mass of a walker as 55kg. It then says she starts from rest and walks 20m is 7s. It wants to know the horizontal force acting on her. 

From kinematics for constant acceleration, I know $\vec{r}=\frac{a}{2}t^2\hat{i}$. Plugging in the known time and the known distance I solved for the acceleration and then I could get the force by multiplying the acceleration by the walker's mass. So I got the problem right... but then I got to wondering: Was there a way to do this problem using energy? I have in mind $\vec{F}\cdot\Delta\vec{r}=\Delta K$. I tried but I don't know the final velocity (from the given information).
Edit: I realized after looking at some of the feedback that I do know the final velocity (because the linear dependance of velocity on time means the average velocity must be half the final velocity). Therefore, you can see below, that I have posted the answer I was hoping to write back when I wished I knew the final velocity.
 A: Assuming constant acceleration from rest the velocity against time graph looks like this:  

Knowing the displacement $s$, which is the area under the graph, and the time $t$ one can link these two quantities either to the acceleration $a$ using $s = \frac 12 \,at\,t = \frac 12at^2$ (compare with the constant acceleration kinematic equation $s = ut + \frac 12 at^2$ with the initial velocity $u = 0$) or the final velocity $v$ using $s = \frac12 \,v\,t$ (compare with the constant acceleration kinematic equation $s = \frac 12 \frac{(u+v)}{t}$ with the initial velocity $u=0$).
One can then use either Newton's second law $F = ma$ or the work-energy theorem $Fs = \frac 12 m v^2$ to find the force $F$.
A: OK, so a female point mass $m$ accelerates from $v=0$ at constant acceleration and covers distance $r$ in time $t$, so using:
$$ d = \frac 1 2 a t^2 $$
we get
$$ a = 2d/t^2 $$
so that:
$$ F = ma = 2md/t^2 $$.
The question is, can this problem be solved using energy? Let's try:
We have to tilt it and use an equivalent gravitational field $a$, in which an at rest mass falls $d$ in time $t$, which mean the potential energy:
$$ U = mad $$
is converted into kinetic energy:
$$ K = ? $$.
Now what? Well, we know the average velocity is:
$$ \bar v = d/t $$
and we know the final velocity is twice the average velocity, so:
$$ v = 2\bar v = 2d/t $$
so that the kinetic energy is:
$$ K = \frac 1 2 m v^2 = 2md^2/t^2 $$
an of course:
$$ K = U $$
so that:
$$ 2md^2/t^2 = mad $$
or:
$$ a = 2d/t^2 $$
Now at this point we could use $F=ma$ and get the right answer, but we're not using Newton's Laws. We're going to use:
$$ F = \frac{\partial U}{\partial d} $$
so plugging $a$ in to the expression for $U$:
$$ U = mad = m(2d/t^2)d = 2md^2/t^2 $$
so 
$$ \partial U/\partial d = 2md/t^2 = F $$
which is correct. So the answer to your question is "yes", you can use energy.
A: Using the work-kinetic energy theorem like you stated is a good start. As you said, that method requires knowing the final velocity. So, just use the basic kinematic relation,
$$ v_{f}^{2} = v_{i}^{2} + 2a\Delta x = 2a\Delta x$$
where $\Delta x$ is the displacement which is given in the problem statement. I think it's kinda straight forward from here:
$$ W = \Delta K $$
$$ F \Delta x = \frac{1}{2}m v_{f}^{2} = \frac{1}{2}m (2a \Delta x) = ma \Delta x$$
$$ F = ma $$
So indeed, Newton's second law is recovered, and you would just use the relation that you provided to find the acceleration. In this problem, using energy involves a bit more work than what you did originally, but it's still a workable path :)
A: So from energy conservation $F.s = mv^2/2$ ;$F.s=ma^2t^2/2$ ; $ F.s=\frac{ 2m(at^2/2)^2}{t^2}$ ;F.s=$ \frac{2m \times (at^2/2)^2}{t^2}= 2ms^2/t^2$ ; note that $v = at$ and $s=at^2/2$ s= displacement v= velocity. I get the force as $F= 2 \times m \times s/t^2$ so i conclude the result can be also obtained by energy conservation. 
A: It occurred to me that since $\vec{v}=at\hat{i}$, it is clear that $v_{final}=2v_{average}$. Well, since $v_{average}=\frac{|\Delta\vec{x}|}{t_{total}}$, we know that $v_{final}=2v_{average}=\frac{2|\Delta\vec{x}|}{t_{total}}$. This means that
$$\vec{F}\cdot\Delta\vec{x}=\Delta K=\frac{1}{2}mv_{final}^2$$
can be solved for $|\vec{F}|$ using the known mass, the known distance, and $\vec{v}_{final}=\frac{2|\Delta\vec{x}|}{t_{total}}$:
$$|\vec{F}|=\biggl(\frac{1}{|\Delta\vec{x}|}\biggr)\biggl(\frac{1}{2}\biggr)m\biggl(\frac{2|\Delta\vec{x}|}{t_{total}}\biggr)^2$$
Note that $\vec{F}$ and $\Delta \vec{x}$ both only have components in the positive $\hat{i}$ direction, so I took for granted that: $$\vec{F}\cdot\Delta\vec{x}=|\vec{F}||\Delta\vec{x}|$$
