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Standard XII

Physics

Applying NLM along 3 Axis

If μ = coef...

Question

If $μ =$ coefficient of kinetic friction between all contact surfaces of the blocks of mass $m_{1}$ and $m_{2}$ . Then, the minimum force required to slide the block of mass $m_{2}$ is?

(2 m_{1} + m_{2}) μ g

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(m_{1} + m_{2}) μ g

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(μ m_{1} + m_{2}) g

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(m_{1} - m_{2}) g

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Solution

The correct option is A $(2 m_{1} + m_{2}) μ g$
For block of mass $m_{1}$
$N_{1} = m_{1} g$
$\Rightarrow T = f_{1} = μ N_{1}$
$= T = μ m_{1} g$
For block of mass $m_{2}$
$m_{2} g + N_{1} = N_{2}$
$\Rightarrow N_{2} = (m_{1} + m_{2}) g$
The minimum force required to slide the block is
$F = f_{1} + f_{2} = μ (N_{1} + N_{2})$
$= μ [m_{1} g + (m_{1} + m_{2}) g]$
So, $F = μ [2 m_{1} + m_{2}] g$ .

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If μ= coefficient of kinetic friction between all contact surfaces of the blocks of mass m1 and m2. Then, the minimum force required to slide the block of mass m2 is?

The correct option is A (2m1+m2)μgFor block of mass m1N1=m1g⇒T=f1=μN1=T=μm1gFor block of mass m2m2g+N1=N2⇒N2=(m1+m2)gThe minimum force required to slide the block isF=f1+f2=μ(N1+N2)=μ[m1g+(m1+m2)g]So, F=μ[2m1+m2]g.

If $μ =$ coefficient of kinetic friction between all contact surfaces of the blocks of mass $m_{1}$ and $m_{2}$ . Then, the minimum force required to slide the block of mass $m_{2}$ is?