A block of mass 5 kg is suspended by a light | vedclass

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(B) Given:Mass $m = 5 \ kg$Acceleration $a = 2 \ m/s^2$Time $t = 3 \ s$Initial velocity $u = 0$Acceleration due to gravity $g = 10 \ m/s^2$Step $1$: C

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$540$

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(B) Given:Mass $m = 5 \ kg$Acceleration $a = 2 \ m/s^2$Time $t = 3 \ s$Initial velocity $u = 0$Acceleration due to gravity $g = 10 \ m/s^2$Step $1$: Calculate the tension $T$ in the thread.Since the elevator is accelerating upward,the tension $T$ is given by:$T = m(g + a)$$T = 5(10 + 2) = 5(12) = 60 \ N$Step $2$: Calculate the displacement $S$ of the block in $3 \ s$.Using the equation of motion $S = ut + \frac{1}{2}at^2$:$S = 0(3) + \frac{1}{2}(2)(3)^2$$S = 0 + 1(9) = 9 \ m$Step $3$: Calculate the work done by tension.Work $W = T \cdot S \cdot \cos(	heta)$Since the tension acts in the direction of displacement,$	heta = 0^{\circ}$,so $\cos(0^{\circ}) = 1$.$W = (60 \ N)(9 \ m)(1) = 540 \ J$

$A$ block of mass $5 \ kg$ is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration $2 \ m/s^2$. The work done by tension on the block during the first $3 \ s$ is $(u = 0)$. (in $J$)

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