Question

Two particles of masses m and 2m are placed on
a smooth horizontal table. A string, which joins
them hangs over the edge supporting a light pulley,
which carries a mass 3m. The two parts of the
string on the table are parallel and perpendicular
to the edge of the table. The parts of the string
outside the table are vertical. Find the acceleration
of the particle of mass 3m.​

Answer 1

Answer:

CORRECT ANSWER IS 9g/17

Explanation:

Explanation is in the image attached.

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Answer 2

Figure

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration\:of\:3m\:mass=\frac{9}{17}g}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Masses \: placed \: on \: table = m \: and \: 2m \\ \\ \tt: \implies Hanged \: mass = 3m \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Acceleration \: of \: 3m \: mass = ?$

• According to given question :

• Let T be the tension in the string, a be the acceleration of mass 2m and 2a be the acceleration of mass a.

$\tt: \implies T = m \times 2a - - - - - (1)$

• The mass 3m will come down with an acceleration:

$\tt: \implies \frac{a + 2a}{2} = \frac{3a}{2} \\ \\ \bold{As \: according \: to \: given \: F.B.D} \\ \tt: \implies 3mg - 2T= 3m \times \frac{3a}{2} \\ \\ \text{Putting \: value \: of \: T }\\ \tt: \implies 3mg - 4ma = \frac{9ma}{2} \\ \\ \tt: \implies 3mg \times 2= 9ma + 8ma \\ \\ \tt: \implies 6mg = 17ma \\ \\ \tt: \implies a = \frac{6mg}{17m} \\ \\ \green{\tt: \implies a = \frac{6g}{17} } \\ \\ \tt{Putting \: value \: of \:( a )\: in \: acc^{n} \: of \:( 3m )\: block} \\ \tt: \implies Acceleration \: of \: 3m \: mass = \frac{3}{2} \times \frac{6g}{17} \\ \\ \green{\tt: \implies Acceleration \: of \: 3m \: mass = \frac{9}{17}g}$