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Standard XII

Physics

Block on a Block Problems

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 the horizontal surface is $0.15$ . Find the minimum weight $m$ that should be put on top of $m_{2}$ to stop the motion.
( $T$ is tension in the string and take $g = 10 m / s^{2}$ )

18.3 k g

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23.3 k g

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43.3 k g

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10.3 k g

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Solution

The correct option is B $23.3 k g$
Given: $m_{1} = 5 k g, m_{2} = 10 k g, μ = 0.15$

For $m_{1}$ :
$\begin{matrix} m_{1} g - T = m_{1} a \\ (1) \end{matrix}$
$\Rightarrow 50 - T = 5 \times a$
For $m_{2}$ :
$T - μ (m_{2} + m) g = (m_{2} + m) a$
$\begin{matrix} T - 0.15 (10 + m) g = (10 + m) a \\ (2) \end{matrix}$

When the motion stops,
$a = 0$ ,
From $(1)$ , $\Rightarrow T = 50$
Substituting the value of $T$ in $(2)$ ,
$50 = 0.15 (m + 10) 10$
$\Rightarrow 5 = \frac{3}{20} (m + 10)$
$\Rightarrow \frac{100}{3} = m + 10$
$∴ m = 23.3 k g$

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The correct option is B 23.3 kgGiven: m1=5 kg,m2=10 kg,μ=0.15 For m1: m1g−T=m1a(1) ⇒50−T=5×a For m2: T−μ(m2+m)g=(m2+m)a T−0.15(10+m)g=(10+m)a(2) When the motion stops, a=0, From (1), ⇒T=50 Substituting the value of T in (2), 50=0.15(m+10)10 ⇒5=320(m+10) ⇒1003=m+10 ∴m=23.3 kg