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Standard XII

Physics

Work Done as Dot Product

Assuming coef...

Question

Assuming coefficient of friction $0.25$ between block and incline. Find the acceleration of each block in the figure.

\frac{g}{5}

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\frac{g}{7}

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\frac{g}{6}

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none of these

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Solution

The correct option is A $\frac{g}{5}$
For $m_{1} : m_{1} a = m_{1} g - T_{1} \Rightarrow 2 a = 2 g - T_{1} . . . . (1)$
For $m_{2} : m_{2} a = T_{1} - T_{2} - m_{2} g sin 37 - μ m g c o s 37$
or. $a = T_{1} - T_{2} - g sin 37 - 0.25 g c o s 37$
$\Rightarrow a = T_{1} - T_{2} - 0.8 g . . . (2)$
For $m_{3} : m_{3} a = T_{2} - m_{3} g \Rightarrow 0.5 a = T_{2} - 0.5 g . . . . (3)$
$(1) + (2) + (3), 3.5 a = 0.7 g \Rightarrow a = \frac{0.7 g}{3.5} = \frac{g}{5}$

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Assuming coefficient of friction 0.25 between block and incline. Find the acceleration of each block in the figure.

The correct option is A g5For m1:m1a=m1g−T1⇒2a=2g−T1....(1)For m2:m2a=T1−T2−m2gsin37−μmgcos37or.a=T1−T2−gsin37−0.25gcos37⇒a=T1−T2−0.8g...(2)For m3:m3a=T2−m3g⇒0.5a=T2−0.5g....(3)(1)+(2)+(3),3.5a=0.7g⇒a=0.7g3.5=g5

Assuming coefficient of friction $0.25$ between block and incline. Find the acceleration of each block in the figure.