Three blocks A,B,and C of masses 5 text{ kg}, | vedclass

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(A) The total mass of the system is $M = m_A + m_B + m_C = 5 	ext{ kg} + 3 	ext{ kg} + 2 	ext{ kg} = 10 	ext{ kg}$.Since the surface is smooth,the

Vedclass · Accepted Answer

$40 	ext{ N}, 64 	ext{ N}$

Vedclass · Answer

(A) The total mass of the system is $M = m_A + m_B + m_C = 5 	ext{ kg} + 3 	ext{ kg} + 2 	ext{ kg} = 10 	ext{ kg}$.Since the surface is smooth,the acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{80 	ext{ N}}{10 	ext{ kg}} = 8 	ext{ m/s}^2$.For block $A$ $(5 	ext{ kg})$,the only horizontal force acting is the tension $T_1$. Therefore,$T_1 = m_A 	imes a = 5 	ext{ kg} 	imes 8 	ext{ m/s}^2 = 40 	ext{ N}$.For block $B$ $(3 	ext{ kg})$,the forces acting are $T_2$ (pulling forward) and $T_1$ (pulling backward). Therefore,$T_2 - T_1 = m_B 	imes a$.Substituting the values,$T_2 - 40 	ext{ N} = 3 	ext{ kg} 	imes 8 	ext{ m/s}^2 = 24 	ext{ N}$.Thus,$T_2 = 40 	ext{ N} + 24 	ext{ N} = 64 	ext{ N}$.Therefore,the tensions $T_1$ and $T_2$ are $40 	ext{ N}$ and $64 	ext{ N}$ respectively.

Three blocks $A$,$B$,and $C$ of masses $5 \text{ kg}$,$3 \text{ kg}$,and $2 \text{ kg}$ respectively are pulled on a horizontal smooth surface by a force of $80 \text{ N}$ as shown in the figure. The tensions $T_1$ and $T_2$ in the strings are respectively:

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