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Standard XII

Physics

Distance

A system of t...

Question

A system of three blocks are connected by strings as shown in figure. Calculate acceleration of each block and tension in the strings.

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Solution

Common acceleration of the system is,
$6 g - g = (6 + 3 + 1) a$
$a = \frac{60 - 10}{6 + 3 + 1}$
$a = 5 m s^{2}$
As they are connected to each other so, all blocks will have common acceleration.
Now,
For, $T_{2}$

$T_{2} - 10 = 5$
$T_{2} = 15 N$
For $T_{1}$

$60 - T_{1} = 30$
$T_{1} = 30 N$

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A system of three blocks are connected by strings as shown in figure. Calculate acceleration of each block and tension in the strings.

Common acceleration of the system is,
$6 g - g = (6 + 3 + 1) a$
$a = \frac{60 - 10}{6 + 3 + 1}$
$a = 5 m s^{2}$
As they are connected to each other so, all blocks will have common acceleration.
Now,
For, $T_{2}$

$T_{2} - 10 = 5$
$T_{2} = 15 N$
For $T_{1}$

$60 - T_{1} = 30$
$T_{1} = 30 N$

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A system of three blocks are connected by strings as shown in figure. Calculate acceleration of each block and tension in the strings.

Common acceleration of the system is,6g−g=(6+3+1)aa=60−106+3+1a=5ms2As they are connected to each other so, all blocks will have common acceleration.Now,For, T2T2−10=5T2=15NFor T160−T1=30T1=30N

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Common acceleration of the system is,
$6 g - g = (6 + 3 + 1) a$
$a = \frac{60 - 10}{6 + 3 + 1}$
$a = 5 m s^{2}$
As they are connected to each other so, all blocks will have common acceleration.
Now,
For, $T_{2}$

$T_{2} - 10 = 5$
$T_{2} = 15 N$
For $T_{1}$

$60 - T_{1} = 30$
$T_{1} = 30 N$