A pendulum with a bob of mass m is suspended from a horizontal platform. The platform is given a horizontal uniform acceleration. The breaking tension in the light string of the pendulum is( 2/√3 ) mg. Find the work done by the extreme tension T on the bob in the first one sec.

Work done, W = (F) (S).    Force F and displacement  are parallel. Here, the bob does not move in y axis.      ⇒ The work done by the tension T along Y axis is zero.    Since the bob moves horizontally (along x-axis) with an acceleration, say a, SFx= ma⇒ Tsin θ= ma⇒ a = T sin θ /m .............. (1)For equilibrium of the bob along Y - axisSFy = 0 ⇒ T cos θ = mg .......... (2)From (1) and (2) ,⇒ a = g tan θ ................. (3)Since the breaking strength .T = 2mg / √3 ⇒ mg / cos θ = 2mg / √3⇒ θ = 300 , Putting θ = 300 in (3)We obtain a = g tan 300 = g / √3Hence required work done = W = F.S.=(T sin θ) (1/2 at2)Where T = 2√3 mg ,θ = 300 , t = 1 sW = (2mg / √3) sin 300 . (1/2 g/√3 . 12)= mg2 / 6