# mechanics question

## mechanics question

1) a body is acted on by a force towards a point. the magnitude of the body is inversely proportional to the square of distance. the path of the body will be?

**rahul**- Posts : 1

Join date : 2008-05-21

## Re: mechanics question

F = k/x^2

where k is a constant of proportionality

F= k/x^2 = mvdv/dx

k/x^2 =mvdv/dx

k/x^2dx = mvdv

on integartion u get

v^2 =k/m(-1/x)

No we need x as

dx/dt = sqrt(k/m * 1/x)

sqrt(m/k) sqrt(x)/dx = dt

2/3x^3/2sqrt(m/k) = t

now in principle we can write x as a function of time

Now we have

x^3/2 = 3/2sqrt(k/m)*t

on simplifying

x = {3/2 sqrt (k/m)*t}^2/3

now we have x is proportional to t^2/3

That would mean that we can describe the x in terms of a power function

You can easily check the function

x =t^2/3

That would represent the path of x

caution it is defined only for x<0

----Hope that helps

where k is a constant of proportionality

F= k/x^2 = mvdv/dx

k/x^2 =mvdv/dx

k/x^2dx = mvdv

on integartion u get

v^2 =k/m(-1/x)

No we need x as

dx/dt = sqrt(k/m * 1/x)

sqrt(m/k) sqrt(x)/dx = dt

2/3x^3/2sqrt(m/k) = t

now in principle we can write x as a function of time

Now we have

x^3/2 = 3/2sqrt(k/m)*t

on simplifying

x = {3/2 sqrt (k/m)*t}^2/3

now we have x is proportional to t^2/3

That would mean that we can describe the x in terms of a power function

You can easily check the function

x =t^2/3

That would represent the path of x

caution it is defined only for x<0

----Hope that helps

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

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