Two masses m 1=10 kg and m 2=20 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 0.30. The minimum weight m that should be put on top of m2 to stop the motion is kg

Question

Two masses m 1=10 kg and m 2=20 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f8887363c171918b432f5add2ab37ed7-.png" /> The coefficient of friction of horizontal surface is 0.30. The minimum weight m that should be put on top of m2 to stop the motion is kg (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fb01269350a363af479dc51cddde70af-.png" /> At equilibrium, F = T μ( m 2+ m ) g = m 1 g ∴ μ(20+ m ) g =10 g ∴ 20+ m =(10/0.30) ∴ 20+ m =(100/3) ∴ m =(40/3)=13.33 kg

Q. Two masses m1​=10kg and m2​=20kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.30. The minimum weight m that should be put on top of m2​ to stop the motion is___________ kg

At equilibrium, F=T μ(m2​+m)g=m1​g ∴μ(20+m)g=10g ∴20+m=0.3010​ ∴20+m=3100​ ∴m=340​=13.33kg

At equilibrium, $F = T$
$μ (m_{2} + m) g = m_{1} g$
$∴ μ (20 + m) g = 10 g$
$∴ 20 + m = \frac{10}{0.30}$
$∴ 20 + m = \frac{100}{3}$
$∴ m = \frac{40}{3} = 13.33 k g$