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 -

Question

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 - (A) 18.3 kg (B) 27.3 kg (C) 43.3 kg (D) 10.3 kg

Tardigrade Admin · Accepted Answer

To stop the moving block m2, acceleration of m2 should be opposite to velocity of m2 m1 g<μ(m+m2) g ⇒ 5<0.15(10+m2) ⇒ m2>23.33 kg ∴ Minimum mass =27.3 kg (according to given options)

Q. 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 -

18.3 kg

27.3 kg

43.3 kg

10.3 kg

To stop the moving block $m_{2}$ , acceleration of $m_{2}$ should be opposite to velocity of $m_{2}$
$m_{1} g < μ (m + m_{2}) g$
$\Rightarrow 5 < 0.15 (10 + m_{2})$
$\Rightarrow m_{2} > 23.33 k g$
$∴$ Minimum mass $= 27.3 k g$ (according to given options)

Q. Two masses m1​=5kg and m2​=10kg, 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 -