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Standard XII

Physics

Static Friction

Two masses m1...

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 kg

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10.3 kg

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18.3 kg

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23.3 kg

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Solution

The correct option is D 23.3 kg
$μ (m + m_{2}) = m_{1}$
$m + m_{2} = \frac{m_{1}}{μ} \Rightarrow m = \frac{m_{1}}{μ} - m_{2}$
$m = \frac{5}{0.15} - 10 = 23.33 k g$

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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 23.3 kgμ(m+m2)=m1 m+m2=m1μ⇒m=m1μ−m2 m=50.15−10=23.33 kg

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 23.3 kg
$μ (m + m_{2}) = m_{1}$
$m + m_{2} = \frac{m_{1}}{μ} \Rightarrow m = \frac{m_{1}}{μ} - m_{2}$
$m = \frac{5}{0.15} - 10 = 23.33 k g$