Newton's laws vs energy for solving a problem











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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 the 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).










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  • 1




    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
    4 hours ago






  • 1




    When I walk, I don’t accelerate uniformly and go faster and faster.
    – G. Smith
    4 hours ago












  • @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
    4 hours ago















up vote
1
down vote

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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 the 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).










share|cite|improve this question




















  • 1




    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
    4 hours ago






  • 1




    When I walk, I don’t accelerate uniformly and go faster and faster.
    – G. Smith
    4 hours ago












  • @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
    4 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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 the 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).










share|cite|improve this question















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 the 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).







homework-and-exercises newtonian-mechanics energy






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edited 3 hours ago

























asked 5 hours ago









okcapp

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  • 1




    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
    4 hours ago






  • 1




    When I walk, I don’t accelerate uniformly and go faster and faster.
    – G. Smith
    4 hours ago












  • @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
    4 hours ago














  • 1




    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
    4 hours ago






  • 1




    When I walk, I don’t accelerate uniformly and go faster and faster.
    – G. Smith
    4 hours ago












  • @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
    4 hours ago








1




1




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
4 hours ago




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
4 hours ago




1




1




When I walk, I don’t accelerate uniformly and go faster and faster.
– G. Smith
4 hours ago






When I walk, I don’t accelerate uniformly and go faster and faster.
– G. Smith
4 hours ago














@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
4 hours ago




@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
4 hours ago










4 Answers
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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.






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    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 :)






    share|cite|improve this answer























    • 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
      4 hours ago










    • 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
      4 hours ago


















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    0
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    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.






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      It occurred to me that since $vec{v}=athat{i}$, it must be true that $v_{final}=2v_{average}$. 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 twice the average speed:
      $$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
      Note that since $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$






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        4 Answers
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        4 Answers
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        up vote
        2
        down vote













        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.






        share|cite|improve this answer

























          up vote
          2
          down vote













          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.






          share|cite|improve this answer























            up vote
            2
            down vote










            up vote
            2
            down vote









            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            JEB

            5,2921717




            5,2921717






















                up vote
                1
                down vote













                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 :)






                share|cite|improve this answer























                • 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
                  4 hours ago










                • 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
                  4 hours ago















                up vote
                1
                down vote













                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 :)






                share|cite|improve this answer























                • 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
                  4 hours ago










                • 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
                  4 hours ago













                up vote
                1
                down vote










                up vote
                1
                down vote









                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 :)






                share|cite|improve this answer














                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 :)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 4 hours ago

























                answered 4 hours ago









                N. Steinle

                907110




                907110












                • 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
                  4 hours ago










                • 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
                  4 hours ago


















                • 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
                  4 hours ago










                • 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
                  4 hours ago
















                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
                4 hours ago




                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
                4 hours ago












                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
                4 hours ago




                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
                4 hours ago










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                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.






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                  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.






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                    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.






                    share|cite|improve this answer














                    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.







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                    edited 1 hour ago

























                    answered 4 hours ago









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                        It occurred to me that since $vec{v}=athat{i}$, it must be true that $v_{final}=2v_{average}$. 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 twice the average speed:
                        $$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
                        Note that since $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$






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                          It occurred to me that since $vec{v}=athat{i}$, it must be true that $v_{final}=2v_{average}$. 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 twice the average speed:
                          $$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
                          Note that since $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$






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                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            It occurred to me that since $vec{v}=athat{i}$, it must be true that $v_{final}=2v_{average}$. 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 twice the average speed:
                            $$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
                            Note that since $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$






                            share|cite|improve this answer














                            It occurred to me that since $vec{v}=athat{i}$, it must be true that $v_{final}=2v_{average}$. 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 twice the average speed:
                            $$|vec{F}|=biggl(frac{1}{|Deltavec{x}|}biggr)biggl(frac{1}{2}biggr)mbiggl(frac{2|Deltavec{x}|}{t_{total}}biggr)^2$$
                            Note that since $vec{F}$ and $Delta vec{x}$ both only have components in the positive $hat{i}$ direction, $$vec{F}cdotDeltavec{x}=|vec{F}||Deltavec{x}|$$







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                            edited 30 mins ago

























                            answered 3 hours ago









                            okcapp

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