In the given figure- the coefficient of friction between m1 and m2 is - and m2 and horizontal surface is zero- Choose the correct option-

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Standard XII

Physics

Block on a Block Problems

In the given ...

Question

In the given figure, the coefficient of friction between $m_{1}$ and $m_{2}$ is $μ$ and $m_{2}$ and horizontal surface is zero. Choose the correct option.

F > μ m_{1} g

, then relative acceleration is found between

m_{1}

and

m_{2}

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F < μ m_{1} g

, then no relative acceleration is found between

m_{1}

and

m_{2}

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F > μ m_{1} g

, then both bodies move togther

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(a) and (b) are correct

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Solution

The correct option is B If $F < μ m_{1} g$ , then no relative acceleration is found between $m_{1}$ and $m_{2}$
The FBDs of the blocks are as shown

Let us consider that there is no relative acceleration between $m_{1}$ and $m_{2}$
$∴$ From FBD we get
$F - f = m_{1} a$ (on mass $m_{1}$ ) ... (1)
$f = m_{2} a$ (on mass $m_{2}$ ) ... (2)
$f = μ_{s} m_{1} g$ $(∵ f = μ_{s} N_{1})$ ... (3)
$\Rightarrow F = (m_{1} + m_{2}) a$

From (1), (2) and (3)
$\Rightarrow F = (m_{1} + m_{2}) \frac{μ_{s} m_{1} g}{m_{2}}$
$F = μ_{s} m_{1} g (1 + \frac{m_{1}}{m_{2}})$
This is maximum possible force so that there is no relative acceleration between $m_{1}$ and $m_{2}$
Or, If $F < μ m_{1} g$ , then no relative acceleration is found between $m_{1}$ and $m_{2}$

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Similar questions

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Q. For the figure shown below, masses

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choose the correct option(s)

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In the given figure, the coefficient of friction between m1 and m2 is μ and m2 and horizontal surface is zero. Choose the correct option.

In the given figure, the coefficient of friction between $m_{1}$ and $m_{2}$ is $μ$ and $m_{2}$ and horizontal surface is zero. Choose the correct option.