A body of mass 1, kg is moving in a vertical | vedclass

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(A) Let the mass of the body be $m = 1\, kg$ and the radius of the circular path be $r = 1\, m$. Let $g = 10\, m/s^2$.At the lowest point,the potentia

Vedclass · Accepted Answer

$20$

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(A) Let the mass of the body be $m = 1\, kg$ and the radius of the circular path be $r = 1\, m$. Let $g = 10\, m/s^2$.At the lowest point,the potential energy is $0$ and at the highest point,the potential energy is $mg(2r) = 2mgr$.By the law of conservation of energy,the total energy at the lowest point equals the total energy at the highest point: $KE_{low} + PE_{low} = KE_{high} + PE_{high}$.$KE_{low} + 0 = KE_{high} + 2mgr$.Therefore,the difference in kinetic energies is $KE_{low} - KE_{high} = 2mgr$.Substituting the values: $2 	imes 1\, kg 	imes 10\, m/s^2 	imes 1\, m = 20\, J$.

$A$ body of mass $1\, kg$ is moving in a vertical circular path of radius $1\, m$. The difference between the kinetic energies at its highest and lowest position is ......... $J$.

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