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Standard XII

Physics

Block on a Block Problems

Two masses ...

Question

Two masses $m_{2} = 10 k g$ and $m_{1} = 5 k g$ are connected by a string passing over a pulley as shown. If the coefficient of friction be 0.15, then the minimum weight that may be placed on $m_{2}$ to stop motion is

18.7 kg

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23.3 kg

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32.5 kg

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44.3 kg

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Solution

The correct option is A 23.3 kg
For equilibrium $(f_{s})_{m a x} = T$ ..........(i)
From figure $T = m_{1} g . . . . . . . . . . (i i)$
and $N = (m_{2} + m) g . . . . . . . . . . . (i i i)$
Also $(f_{s})_{m a x} = μ_{s} N . . . . . . . . . . . (i v)$
$\Rightarrow μ_{s} (m_{2} + m) g = m_{1} g$
or $m = \frac{m_{1}}{μ_{s}} - m_{2} = \frac{5}{0.15} - 10 \Rightarrow m = 23.3 k g$

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Two masses m2=10kg and m1=5kg are connected by a string passing over a pulley as shown. If the coefficient of friction be 0.15, then the minimum weight that may be placed on m2 to stop motion is

The correct option is A 23.3 kgFor equilibrium (fs)max=T..........(i)From figure T=m1g..........(ii)and N=(m2+m)g...........(iii)Also (fs)max=μsN...........(iv)⇒μs(m2+m)g=m1gor m=m1μs−m2=50.15−10⇒m=23.3kg

Two masses $m_{2} = 10 k g$ and $m_{1} = 5 k g$ are connected by a string passing over a pulley as shown. If the coefficient of friction be 0.15, then the minimum weight that may be placed on $m_{2}$ to stop motion is