Forces acting on a particle moving in a straight line varies with the velocity of the particle as `F = (alpha)/(upsilon)` where `alpha` is constant. The work done by this force in time interval `Delta t` is :

Options

A. `alpha Delta t`
B. `(1)/(2) alpha Delta t`
C. `2 alpha Delta t`
D. `alpha^(2) Delta t`

Answer

A

Explanation

`F=(alpha)/(upsilon) rArr m(d upsilon)/(dt)=(alpha)/(upsilon)rArr int m upsilon d upsilon = int alphadt`
`((m upsilon^(2))/(2)) = alpha t , Delta KE = alpha Delta r =` work done.