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QJEEM18Phf7d4: JEE Main 2018 Physics - Class 11 JEEM18Phf7d4 Solution: Two masses m 1 = 5 kg and m 2 = 10 kg, connected by an…

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 : JEE Main 2018 (Offline) Physics - Laws of Motion Question 111 English

Multiple Choice Options

  • A 27.3 kg
  • B 43.3 kg
  • C 10.3 kg
  • D 18.3 kg

Step-by-Step Verified Solution

Moving block will stop when the friction force between m2 and surface is ≥\ge tension force.

So condition for stopping the moving block,

f≥Tf \ge T

⇒μN≥T\Rightarrow \mu N \ge T

⇒μ(m+m2)g≥m1g\Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g

When m is minimum then,

μ(m+m2)g=m1g\mu \left( {m + {m_2}} \right)g = {m_1}g

⇒m=m1−μm2μ\Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }

⇒m=5−0.15×100.15\Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}} = 23.33 kg

So if m ≥\ge 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg.

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