# 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.

• This problem is extremely unclear (not your fault). What is a walker? Is it a person or a thing? Is there friction? If we're talking about a human walking, then that sounds like common-core, because the kinesiology of walking is not amenable to simple analysis, which is why physics problems general discuss masses on frictionless surfaces. – JEB Nov 24 '18 at 1:14
• When I walk, I don’t accelerate uniformly and go faster and faster. – G. Smith Nov 24 '18 at 1:16
• @JEB please take the force of static friction of ground on walker to be the only relevant force; and treat the walker as a point mass. – okcapp Nov 24 '18 at 1:17

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 $$F$$.

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.

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 :)

• I'm a little confused by this. Your kinematic equation is exactly the same as your second equation, so you've somehow used the same equation twice to recover Newton's second law. I"m not sure what exactly you did. There is a typo in the last equation, by the way. – garyp Nov 24 '18 at 1:45
• Thank you for pointing out the typo, it's been fixed. And to clarify, I agree that the kinematic equation I provided can be algebraically manipulated into the work-kinetic energy theorem, but if you want to use energy to solve the OP's problem then you need the final velocity, and if you want to use the work-kinetic energy theorem then it's a rather circular method of solving. One can instead use an artificial potential energy as JEB did in his solution, but I just wanted to point out the circularity of the OP's supposition. – N. Steinle Nov 24 '18 at 1:53

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 $$\vec{v}=at\hat{i}$$, 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: $$\vec{F}\cdot\Delta\vec{x}=|\vec{F}||\Delta\vec{x}|$$