Force acting on a particle moving in a straight line varies with the velocity of the particle as F=k/v,where K is a constant.The work done by this force in time t is:

(A)K/v^2.t

(B)2Kt

(C)Kt

(D)2Kt/v^2

we are given

F = ma = K/v

or

m(dv/dt) = K/v

or

mvdv = Kdt

now integrating both sides

v varies form v_{1} to v_{2} while t from 0 to t, so

m(v_{2}^{2}/2 - v_{1}^{2}/2) = Kt

so,

(1/2)mv_{2}^{2}/2 - (1/2)mv_{1}^{2} = Kt

thus,

change in KE = Kt

now, from work energy theorem

change in KE = work done

thus,

**work done, W = Kt**

**correct answer is (c)**

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