A particle moves in a straight line with retardation proportional to its displacement. Its loss in kinetic energy for any displacement x is proportional to

Options

A. `x^(2)`
B. `e^(x)`
C. x
D. `log_(e)x`

Answer

1

Explanation

As given in question, retardation (negative acceleration)` a prop x`
`rArr " " a=-kx`
Where K is a proportionality constant
`rArr " " vdv=kxdx`
Integrating, we get `int_(v_(i))^(v_(f)) vdv=-int_(0)^(x) kx dx`
Where `v_(i)` and `v_(f)` respectively are intial and final velocities of particle
`((v^(2))/2)_(v_(i))^((v_(f))=-k((x^(2))/2)_(0)^(2)`
`rArr " " (v_(f)^(2))/2-(v_(i)^(2))/2=-k (x^(2))/2`
`rArr " " 1/2 mv_(f)^(2) -1/2 mv_(i)^(2) =(-1)/2 mkx^(2)`
`rArr " " K.E._("final")-K.E._("initial")=1/2 mkx^(2)`
Hence, loss in kinetic energy `prop x^(2)`