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Standard IX

Physics

Second Law of Motion

Two masses ...

Question

Two masses $m_{1} = 5 k g$ and $m_{2} = 10 k g$ , connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$ . The minimum weight m that should be put on top of $m_{2}$ to stop the motion is:

43.3 k g

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10.3 k g

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18.3 k g

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27.3 k g

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Solution

The correct option is D $27.3 k g$
from FBD of mass $m_{1}$ we get $t = 5 g$
to stop the motion $T \leq μ (m_{2} + m) g$
$=> 5 g \leq 0.15 (10 + m) g$
$=> m \geq 23.3 k g$
so minimum mass that can stop motion is 23.3kg but among the options best option is option D.

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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:

The correct option is D 27.3 kgfrom FBD of mass m1 we get t=5gto stop the motion T≤μ(m2+m)g=>5g≤0.15(10+m)g=>m≥23.3kgso minimum mass that can stop motion is 23.3kg but among the options best option is option D.

Two masses $m_{1} = 5 k g$ and $m_{2} = 10 k g$ , connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$ . The minimum weight m that should be put on top of $m_{2}$ to stop the motion is:

The correct option is D $27.3 k g$
from FBD of mass $m_{1}$ we get $t = 5 g$
to stop the motion $T \leq μ (m_{2} + m) g$
$=> 5 g \leq 0.15 (10 + m) g$
$=> m \geq 23.3 k g$
so minimum mass that can stop motion is 23.3kg but among the options best option is option D.