Question

Dear Physics Wizard, please help me make sense of this question.....A mass m moves along x axis due to a single force F=Fxihat. It has constant mechanical energy given by E=1/2mv^2 - c1x + c2x^2 where c1 and c2 are constants. Find expression for Fx

Answer 1

i think that we need to use energy conservation here
there's only one force acting and that too in a uniform direction
energy conservation says that total work done is equal to change in the energy(i think so haven't done it from weeks together)
so we can say
\[W=\Delta E\]
we know that dW=Fdx
so \[Fdx=\Delta E\]
now to find F, differentiate the equation of energy with respect to x and you may put limits from 0 to x
so i got it as F=2(c2)x-c1
hope this helps you i am weak at it

Answer 2

@AllTehMaffs what do you think??

Answer 3

I think you can separate the Energy into kinetic and potential (the kinetic being the only part that's moving (by definition))
\[E_K = \frac{1}{2}mv^2\]
\[E_P=-c_1x+c_2x^2\]
Then work done with the kinetic part is
\[W_1=F_1\Delta x = \Delta E_K\]
\[F_1 = \frac{ \Delta E_k}{\Delta x}\]
and work done on the potential part is
\[W_2 =F_2 \Delta x = -\Delta E_P\]
\[F_2= -\frac{\Delta E_P}{\Delta x}\]
So the net force (in the i direction, since it's constrained in 1D would be

\[F_x = F_1+F_2 = \frac{\Delta E_k}{\Delta x} - \frac{\Delta E_P}{\Delta x}\]
\[F_x = \frac{m v_2^2-mv_1^2}{2 \Delta x} - \frac{-c_1 \Delta x + c_2(x_2^2-x_1^2)}{\Delta x}\]
\[=\frac{m v_2^2-mv_1^2}{2 \Delta x}+c_1-c_2\frac{x_2^2-x_1^2}{\Delta x}\]
\[F_x=\frac{m(v_2^2-v_1^2)+2c_1+2 c_2(x_1^2-x_2^2)}{2\Delta x}\]

Answer 4

i think you are right

Answer 5

I like yours though - I wasn't sure how to fit the differential in, so stuck with deltas

Answer 6

Mine looks awkward somehow though :P

Answer 7

yours is a very good try and seems to be more logical

Answer 8

Thanks ^_^ I'm always terrible at these sorts of problems :P

Answer 9

no man:) you do it very well

Answer 10

oops, missed a term

\[F_x=\frac{m(v_2^2-v_1^2)+2c_1\Delta x+2 c_2(x_1^2-x_2^2)}{2\Delta x}\]

Answer 11

Wish I know how to do it with calc though! ^^

Answer 12

and thanks again - you're kind!

Answer 13

calculus is easy you just need to know the rules any you're on!

Answer 14

It definitely always makes things easier when the problem is set up right ^^ Plus it's prettier!

Answer 15

yup you're right

Answer 16

Wow thank you so much. I can follow everything you are ding but can you please explain in a little more detail why we can break up the initial equation into kinetic and potential energy please (;

Answer 17

Because the definitions of work are different for kinetic and potential energy

For kinetic energy
\[ W=F \Delta x=\Delta K=K_2-K_1 \quad \quad \quad \quad \longrightarrow \quad F_1 = \frac{K_2-K_1}{\Delta x}\]
but for potential
\[W=- \Delta U=-(U_2-U_1)=U_1-U_2 \quad \longrightarrow \quad F_2 = \frac{U_1-U_2}{\Delta x}\]

\[F_x=F1+F2 = \frac{K_2-K_1}{\Delta x} + \frac{U_1-U_2}{\Delta x} = \frac{K_2-K_1 + U_1-U_2}{\Delta x}\]
So if we had just tried to say
\[W= \Delta E=E_2-E_1\]

There would be an incorrect sign on U2 and U1

We would end up with
\[F_x = \frac{E_2-E_1}{\Delta x} = \frac{ K_2 -K_1 + U_2-U_1}{\Delta x}\]