(A) First, consider the system as a whole. The total mass of the system is $(m_1 + m_2)$.Since the surface is smooth, the acceleration $(a)$ of the sy

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$\left(\frac{m_1}{m_1+m_2}\right) F$

(A) First, consider the system as a whole. The total mass of the system is $(m_1 + m_2)$.Since the surface is smooth, the acceleration $(a)$ of the system is given by Newton's second law: $a = \frac{F}{m_1 + m_2}$.Now, consider the free body diagram of mass $m_1$. The only horizontal force acting on $m_1$ is the tension $(T)$ in the string.Applying Newton's second law to mass $m_1$: $T = m_1 \times a$.Substituting the value of acceleration $(a)$: $T = m_1 \times \left(\frac{F}{m_1 + m_2}\right)$.Therefore, the tension in the string is $T = \left(\frac{m_1}{m_1 + m_2}\right) F$.

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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Two masses $m_1$ and $m_2$ are placed on a smooth horizontal surface and are connected by a string of negligible mass. $A$ horizontal force $F$ is applied on the mass $m_2$ as shown in the figure. The tension in the string is

TS EAMCET 2016 All TS EAMCET

A
$\left(\frac{m_1}{m_1+m_2}\right) F$
B
$\frac{m_2 F}{m_1+m_2}$
C
$\left(\frac{m_1}{m_2}\right) F$
D
$\frac{m_2 F}{m_1}$

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Class 11

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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)

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Two masses $m_1$ and $m_2$ are placed on a smooth horizontal surface and are connected by a string of negligible mass. $A$ horizontal force $F$ is applied on the mass $m_2$ as shown in the figure. The tension in the string is

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$A$ block of mass $M$ is lying on a horizontal frictionless surface. One end of a uniform rope of mass $M/2$ is fixed to the block, which is pulled in the horizontal direction by applying a force $F$ at the other end. The tension in the middle of the rope will be

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