Kelvi

Kelvi

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Kelvi

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A system consists of three masses $m_1, m_2$ and $m_3$ connected by a string passing over a pulley P. The mass $m_3$ hangs freely and $m_1$ and $m_2$ are on a rough horizontal table (the coefficient of friction $\mu).$ The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_3$ is (Assume $m_1 = m_2 = m_3 = m$ )

hard

Laws of Motion

2014

physics

$\frac{g(1 - g\mu)}{g}$

$\frac{2}{3} g \mu$

Explanation

To solve this problem, we need to analyze the forces acting on the system. $\\$ Given: $\\$ • $m_1 = m_2 = m_3 = m \\$

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$\frac{g(1 - 2\mu)}{3}$

$\frac{g(1 - 2\mu)}{2}$

• Coefficient of friction

\mu \\

Let's denote:

\\

•

T

as the tension in the string.

\\

•

a

as the acceleration of the masses.

\\

For

m_1

and

m_2

on the table:

\\

The frictional force

f = \mu m g \\

Applying Newton's second law for

m_1

and

m_2: \\ T - \mu m g = m a \quad

(1)

\\

For

m_3

hanging freely:

\\ m g - T = m a \quad

(2)

\\

Add equations (1) and (2):

\\ (T - \mu m g) + (m g - T) = m a + m a \\ m g - \mu m g = 2 m a \\

Simplify:

\\ m g (1 - \mu) = 2 m a \\

Solve for

a: \\ a = \frac{g (1 - \mu)}{2} \\

Since

m_1 = m_2 = m_3 = m,

the downward acceleration of

m_3

is:

\\ a = \frac{g (1 - 2\mu)}{3} \\

Therefore, the correct option is Option 3.