(A) Let $m_1 = 12\,kg$ and $m_2 = 8\,kg$. The acceleration $a = 2.2\,m/s^2$.For the lower mass $(m_2 = 8\,kg)$:The forces acting are tension $T_2$ upw

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(A) Let $m_1 = 12\,kg$ and $m_2 = 8\,kg$. The acceleration $a = 2.2\,m/s^2$.For the lower mass $(m_2 = 8\,kg)$:The forces acting are tension $T_2$ upwards and weight $m_2g$ downwards.$T_2 - m_2g = m_2a$$T_2 - 8(9.8) = 8(2.2)$$T_2 - 78.4 = 17.6$$T_2 = 96\,N$For the system of both masses $(m_1 + m_2 = 20\,kg)$:The forces acting are tension $T_1$ upwards and total weight $(m_1+m_2)g$ downwards.$T_1 - (m_1+m_2)g = (m_1+m_2)a$$T_1 - 20(9.8) = 20(2.2)$$T_1 - 196 = 44$$T_1 = 240\,N$Therefore,$T_1 = 240\,N$ and $T_2 = 96\,N$.

Class 11 Physics Newton's Laws of Motion and Friction Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

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$A$ body of mass $8\,kg$ is hanging from another body of mass $12\,kg$. The combination is being pulled upwards by a string with an acceleration of $2.2\,m/s^2$. The tensions $T_1$ (in the top string) and $T_2$ (in the string between the two masses) will be respectively: (use $g = 9.8\,m/s^2$)

A
$240\,N, 96\,N$
B
$200\,N, 80\,N$
C
$220\,N, 90\,N$
D
$260\,N, 96\,N$

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Newton's Laws of Motion and Friction

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Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

Two blocks of mass $8\,kg$ and $2\,kg$ are connected by a string and they are released on an inclined plane of inclination $30^o$ as shown in the figure. What will be the tension in the string connecting the two blocks (in $,N$)? (Assume the surface is frictionless and $g = 10\,m/s^2$)

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With what acceleration can a block be moved upwards using a rope,if the breaking strength of the rope is $5$ times the weight of the block?

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$A$ uniform rope of mass $1.0\, kg$ is connected to a box of mass $2.0\, kg$,which is placed on a smooth horizontal surface. The free end of the rope is pulled horizontally by a force of $6\, N$. Find the tension at the midpoint of the rope in $N$.

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$A$ man slides down a light rope whose breaking strength is $\eta$ times the weight of the man $(\eta < 1)$. What is the maximum acceleration of the man so that the rope does not break?

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At $t = 0$, a body of mass $100 \text{ g}$ starts moving under the influence of a force $(5\hat{i} + 10\hat{j}) \text{ N}$. After $2 \text{ s}$, its position is $(2x\hat{i} + 5y\hat{j}) \text{ m}$. The ratio $x : y$ is . . . . . . .

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$A$ body of mass $8\,kg$ is hanging from another body of mass $12\,kg$. The combination is being pulled upwards by a string with an acceleration of $2.2\,m/s^2$. The tensions $T_1$ (in the top string) and $T_2$ (in the string between the two masses) will be respectively: (use $g = 9.8\,m/s^2$)

Similar Questions

Two blocks of mass $8\,kg$ and $2\,kg$ are connected by a string and they are released on an inclined plane of inclination $30^o$ as shown in the figure. What will be the tension in the string connecting the two blocks (in $,N$)? (Assume the surface is frictionless and $g = 10\,m/s^2$)

With what acceleration can a block be moved upwards using a rope,if the breaking strength of the rope is $5$ times the weight of the block?

$A$ uniform rope of mass $1.0\, kg$ is connected to a box of mass $2.0\, kg$,which is placed on a smooth horizontal surface. The free end of the rope is pulled horizontally by a force of $6\, N$. Find the tension at the midpoint of the rope in $N$.

$A$ man slides down a light rope whose breaking strength is $\eta$ times the weight of the man $(\eta < 1)$. What is the maximum acceleration of the man so that the rope does not break?

At $t = 0$, a body of mass $100 \text{ g}$ starts moving under the influence of a force $(5\hat{i} + 10\hat{j}) \text{ N}$. After $2 \text{ s}$, its position is $(2x\hat{i} + 5y\hat{j}) \text{ m}$. The ratio $x : y$ is . . . . . . .

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