Cfxtrjtrk

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 $vec{r}=frac{a}{2}t^2hat{i}$. 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 $vec{F}cdotDeltavec{r}=Delta K$. I tried but I don't know the final velocity (from the given information).

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:

$$ v = 2bar v = 2d/t $$

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.

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,

$$ v_{f}^{2} = v_{i}^{2} + 2aDelta x = 2aDelta x$$

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 :)

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 $F$.

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.

It occurred to me that since $vec{v}=athat{i}$, it is clear that $v_{final}=2v_{average}$. Well, since $v_{average}=frac{|Deltavec{x}|}{t_{total}}$, we know that $v_{final}=2v_{average}=frac{2|Deltavec{x}|}{t_{total}}$. This means that
$$vec{F}cdotDeltavec{x}=Delta K=frac{1}{2}mv_{final}^2$$
can be solved for $|vec{F}|$ using the known mass, the known distance, and $vec{v}_{final}=frac{2|Deltavec{x}|}{t_{total}}$:
$$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
Note that $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, so I took for granted that: $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$