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(B) According to the law of conservation of mechanical energy,since the surface is frictionless,the total mechanical energy remains constant.Let $h_A

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$140$

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(B) According to the law of conservation of mechanical energy,since the surface is frictionless,the total mechanical energy remains constant.Let $h_A = 14\, m$ be the height at position $A$ and $h_C = 7\, m$ be the height at position $C$.The mass of the block is $m = 2\, kg$ and acceleration due to gravity is $g = 10\, m/s^2$.The block is released from rest at $A$,so its initial kinetic energy $K_A = 0$.The total energy at $A$ is $E_A = K_A + U_A = 0 + mgh_A = 2 	imes 10 	imes 14 = 280\, J$.The total energy at $C$ is $E_C = K_C + U_C = K_C + mgh_C = K_C + 2 	imes 10 	imes 7 = K_C + 140\, J$.Since $E_A = E_C$,we have $280 = K_C + 140$.Therefore,$K_C = 280 - 140 = 140\, J$.

The figure shows the vertical section of a frictionless surface. $A$ block of mass $2\, kg$ is released from position $A$. Its kinetic energy $(KE)$ as it reaches position $C$ is ................ $J$.

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