A particle moves in a straight line with retardation propotional to its displacement. Loss of K.E. for any displacement x is proportional to :

Options

A. x
B. `x^2`
C. log x
D. `e^x`

Answer

b

Explanation

Here `a=(dv)/(dt)=(dx)/(dt)xx(dv)/(dx)=v(dx)/(dx)`
`v(dv)/(dx)=betax` where `beta` is a constant.
`int_0^v vdv=int_0^x-betaxdx`
or `(v^2-u^2)/(2)=-beta(x^3)/(2)`
or `1/2(mv^2-mu^2)=(-1)/(2)betamx^2` or loss in K.E. is proportional `x^2`