Question

Three blocks of masses ${m_1}$,${m_2}$and ${m_3}$ kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass ${m_1}$, the acceleration of ${m_3}$ will be $a = \dfrac{F}{{{m_1} + {m_2} + {m_3}}}$. So, find the force experienced by ${m_2}$is:
A. $F$
B. $\dfrac{{F \times ({m_1})}}{{{m_2} + {m_3}}}$
C. $\dfrac{{F \times ({m_2} + {m_1})}}{{{m_2} + {m_3}}}$
D. $\dfrac{{F \times ({m_2} + {m_3})}}{{{m_1} + {m_2} + {m_3}}}$

Answer 1

The final acceleration undergone by mass ${m_3}$ is given in the question. Use the force formula for each block to identify the force on block 2 .

Complete step by step answer:
We know that the force experienced by an object is the product of its mass and acceleration. The given setup is placed on a smooth horizontal table, hence there won’t be any friction during acceleration. $F = m \times a$
When the force F is exerted on block 1, since all the blocks are in contact, the acceleration experienced will be the same for all blocks in total, due to a void of friction. Hence a1=a2=a3 .
Now, it is given that the acceleration experienced by block ${m_3}$ is equal to$a = \dfrac{F}{{{m_1} + {m_2} + {m_3}}}$, which is just the ratio of force experienced by the last block by the sum of masses. Since all the blocks are in contact with each other, the acceleration of each block will be the same, however, the same logic doesn’t apply to the force experienced by the block.
Block 2 of mass${m_2}$will experience a force from mass ${m_1}$block when it is exerted by a force. So the total force experienced by the block of mass ${m_2}$ will be the sum of force F and the force exerted by block 1 on 2.
The total force exerted is given as F.
Now the force experienced by block 2 is ,
${F_2} = {m_{(1and2)}} \times a$(Since all 3 are in contact)
$\Rightarrow {F_2} = ({m_1} + {m_2}) \times a$
We know that a is same for all and the value is $a = \dfrac{F}{{{m_1} + {m_2} + {m_3}}}$
Substituting a in the equation, we get
$\Rightarrow {F_2} = ({m_1} + {m_2}) \times \dfrac{F}{{{m_1} + {m_2} + {m_3}}}$
$\Rightarrow {F_2} = \dfrac{{F \times ({m_1} + {m_2})}}{{{m_1} + {m_2} + {m_3}}}$
Hence, Option (b) is the right answer for the given question.

Note: $F = m \times a$ is the mathematical expression of Newton’s second law, which states that the acceleration of an object produced by a net force is directly proportional to the magnitude of the force and inversely proportional to the mass of the object.