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

Options

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

Answer

A

Explanation

`V(dv)/(dx)=-Kx, [(v^(2))/2]_(u)^(v)=-[(kx^(2))/2]_(0)^(x)`
`V^(2)-u^(2)=-Kx^(2)`
`1/2 m u^(2)-1/2 mV^(2)=1/2 mKx^(2)`
Loss `alpha x^(2)`