Consider an elliptically shaped rail PQ in th | vedclass

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(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:$W_{total} = W_{force} + W_{gra

Vedclass · Accepted Answer

$5$

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(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:$W_{total} = W_{force} + W_{gravity} = K_f - K_i$Given that the block starts from rest,$K_i = 0$. The force $F = 18 \ N$ is applied parallel to the displacement vector $\vec{PQ}$. The length of the displacement $PQ$ is calculated using the Pythagorean theorem:$PQ = \sqrt{OP^2 + OQ^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \ m$The work done by the force is:$W_{force} = F 	imes PQ = 18 \ N 	imes 5 \ m = 90 \ J$The work done by gravity is:$W_{gravity} = -mgh = -1 \ kg 	imes 10 \ m/s^2 	imes 4 \ m = -40 \ J$Therefore,the final kinetic energy $K_f$ is:$K_f = 90 \ J - 40 \ J = 50 \ J$We are given $K_f = (n 	imes 10) \ J$,so:$n 	imes 10 = 50 \implies n = 5$

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