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HELP mechanics need my answers checking

please helppppppp :frown:

A particle p of unit mass moves along x axis under influence of a force

f(x)= 72/x^3 - 12/x^2

i.) work out v(x)?
ii.) write down energy conv equation?
iii.) initially p is set in motion with speak square root(3/8) at the point x=8 determine enegry associated with the initial state?
iv.) show that the stationary points are at x= 4 and x= 12
v.) express time taken for particle to travel frmo x=4 to x=12 as a defin integral with respect to x?

i got:
i.) v(x)=-integrate f(x)
36/x^2 - 12/x^-1

ii.) T+V=E

iii.) o.5*(3/8) + (36/64 - 12/8)= -0.75 :confused:

iv.) 0.75=36/x^2 - 12/x
3x^2= 144- 48x
x^2+16x-48=0
(x-4)(x-12)=0 x= 4 and 12

v.) erm no idea HELP?

Reply 1

ukgea

6

iv) I think it should be -0.75 = 36/x^2 - 12/x. There's no problem with negative energies. The potential energy is only relative, so to speak, i.e. what only matters is the difference between potential energy between two points, you could set the actual value of the potential energy to anything you want and it will still work. This can be seen also because the integral

V(x) = \displaystyle \int f(x) dx

generates a + C constant, which you can set to anything you want really.

v.) Consider the small interval from

x

to

x + dx

. What is the particle's speed in this interval? And how long is it? So how long will it take for the particle to pass through?

Reply 2

kanz

OP

ukgea

iv) I think it should be -0.75 = 36/x^2 - 12/x. There's no problem with negative energies. The potential energy is only relative, so to speak, i.e. what only matters is the difference between potential energy between two points, you could set the actual value of the potential energy to anything you want and it will still work. This can be seen also because the integral

V(x) = \displaystyle \int f(x) dx

generates a + C constant, which you can set to anything you want really.

v.) Consider the small interval from

x

to

x + dx

. What is the particle's speed in this interval? And how long is it? So how long will it take for the particle to pass through?

iv>) if i use -0.75 i cant get 4 and 12???

v.) hmm i get spead as square root of 3 and therefore i think its 8/sqr3???

Reply 3

ukgea

6

iv) Yes you can, there's a mistake in your very last line of working.

v) Er, no. I was more meaning for you to work out the speed as a function of position, x. (Hint: You know the kinetic energy as a function of position!)

Reply 4

kanz

OP

ukgea

iv) Yes you can, there's a mistake in your very last line of working.

v) Er, no. I was more meaning for you to work out the speed as a function of position, x. (Hint: You know the kinetic energy as a function of position!)

lol ops i get iv.) :smile:

my mistake

v.) ive tried v. but i still dont get it i need to express it as an integral hmm but i dont know how :confused:

Reply 5

ukgea

6

Okay. No problem. For v), basically, there's this formula:

\displaystyle T = \int_{x_1}^{x_2} \frac{dx}{v}

where T is the time taken for the particle to move from

x_1

to

x_2

, and where v denotes the velocity of the particle (as a function of position x)

(Warning: Slightly dogdy maths below)

The formula can be deduced by the substitution x = x(t), where x(t) is the position of the particle at a time t. Using this substitution, we have dx = v dt, and so

\displaystyle \int_{x_1}^{x_2} \frac{dx}{v}

\displaystyle \int_{t_1}^{x_2} dt

= t_2 - t_1 = T

(where

t_1, t_2

are the times at which the particle is at

x_1, x_2

respectively.

So can you see how you can work out v as a function of position x? And how you can then apply the formula?

Reply 6

v-zero

14

Surely integrating f(x) to get v(x) is wrong, since f(x) = ma and m=1 then surely we get a = 72/x^3 - 12/x^2. In this case we have acceleration in terms of displacement, so the identity dv/dt = (dx/dt)(dv/dx) is the most useful form for the accelleration. This leads to:

v(dv/dx) = 72(x^-3) - 12(x^-2)

=> (v)dv = ( 72(x^-3) - 12(x^-2) )dx

Integrating this will give an equality involving v^2 - then apply a sqaure root to the other side to find v(x).

At least, I think that's what you should be getting at...

Since all the energy is converted to kinetic energy we can find that E = 1/2(m)(v^2) - 1/2(m)(u^2) which simplifies to E = 1/2(v^2 - u^2).

Now, finally let "n" represent the equation found for the velocity of the particle in terms of x, now we can find that:

v = dx/dt = n . This leads to dt = (1/n)dx . Now once again integrate to get t = int (1/n)dx .

Finally use this integral to find the area between x=4 and x=12 as a definite integral.

HELP mechanics need my answers checking

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