A system of 10 balls,each of mass m = 2 ; kg, | vedclass

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(A) When the $6^{th}$ ball just leaves the table,$6$ balls are hanging vertically and $4$ balls are on the horizontal table.The total mass of the syst

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) When the $6^{th}$ ball just leaves the table,$6$ balls are hanging vertically and $4$ balls are on the horizontal table.The total mass of the system is $M = 10m$.The driving force is the weight of the $6$ hanging balls,which is $F = 6mg$.The acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{6mg}{10m} = \frac{3g}{5}$.To find the tension $T$ between the $7^{th}$ and $8^{th}$ ball,we consider the system of the $3$ balls $(8^{th}, 9^{th}, 10^{th})$ remaining on the table.The only horizontal force acting on this system is the tension $T$.Applying Newton's second law to these $3$ balls: $T = (3m)a$.Substituting the value of $a$: $T = 3m 	imes \frac{3g}{5} = \frac{9mg}{5}$.Given $m = 2 \; kg$ and taking $g = 10 \; m/s^2$:$T = \frac{9 	imes 2 	imes 10}{5} = \frac{180}{5} = 36 \; N$.

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