A uniform chain of length 3, m and mass 3, kg | vedclass

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(A) Let the total length of the chain be $L = 3\, m$ and total mass be $M = 3\, kg$. The linear mass density is $\lambda = M/L = 3/3 = 1\, kg/m$.Initi

Vedclass · Accepted Answer

$40$

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(A) Let the total length of the chain be $L = 3\, m$ and total mass be $M = 3\, kg$. The linear mass density is $\lambda = M/L = 3/3 = 1\, kg/m$.Initially,$2\, m$ of the chain is on the table and $1\, m$ is hanging. The center of mass of the hanging part is at a distance of $0.5\, m$ below the table level. Taking the table surface as the reference level $(U = 0)$,the initial potential energy is $U_i = -(m_{hanging})g(h_{cm}) = -(1\, kg)(10\, m/s^2)(0.5\, m) = -5\, J$.The initial kinetic energy is $K_i = 0$.Finally,when the chain completely slips off,the entire chain of length $3\, m$ hangs vertically. Its center of mass is at $1.5\, m$ below the table level. The final potential energy is $U_f = -(M)g(h_{cm}) = -(3\, kg)(10\, m/s^2)(1.5\, m) = -45\, J$.By the law of conservation of mechanical energy,$K_i + U_i = K_f + U_f$.$0 + (-5\, J) = K_f + (-45\, J)$.$K_f = 45 - 5 = 40\, J$.Thus,the value of $k$ is $40$.

$A$ uniform chain of length $3\, m$ and mass $3\, kg$ overhangs a smooth table with $2\, m$ laying on the table. If $k$ is the kinetic energy of the chain in joules as it completely slips off the table,then the value of $k$ is (Take $g = 10\, m/s^2$).

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