A particle move in a straight line with retardation proportional to its displacement its loss of kinectic energy for any displacement `x` is proportional to

Explanation

Here, mass`=`m. Retardation `(-a)` of particle is proportional to displacement `(x)`
`:. -apropx or a=-kx` ..(i)
Now, `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt)=(dv)/(dx)(v)`
From (i), `v(dv)/(dx)=-kx`
or `v dv=-kx dx ` …(ii)
Suppose velocity of particle changes from u to v during its displacement x.
`:.` Intefrating (ii), within proper limits,
`int_(u)^(v)vdv=-kint_(0)^(x)xdx`
`[(v^(2))/(2)]_(u)^(v)=-k[(x^(2))/(2)]_(0)^(x)`
`(1)/(2)v^(2)-(1)/(2)u^(2)=-k(x^(2))/(2)`
or `(1)/(2)u^(2)-(1)/(2)v^(2)=(1)/(2)kx^(2)`
Loss in KE`=(1)/(2)m u^(2)-(1)/(2)mv^(2)=(1)/(2)mkx^(2)`