Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is $\mu_k$ . When the block A is sliding on the table, the tension in the string is:

$\frac{(m_2 + \mu_k m_1)g}{m_1 + m_2}$

$\frac{(m_2 - \mu_k m_1)g}{m_1 + m_2}$

$\frac{m_1 m_2 (1 + \mu_k)g}{m_1 + m_2}$

$\frac{m_1 m_2 (1 - \mu_k)g}{m_1 + m_2}$

Step-by-Step Solution

Free Body Diagram of Block B ( $m_2$ ):

Forces acting are gravity ( $m_2g$ ) downwards and tension ( $T$ ) upwards.
Equation of motion: $m_2g - T = m_2a$ ... (i) (assuming acceleration $a$ downwards).

Free Body Diagram of Block A ( $m_1$ ):

Forces acting horizontally are tension ( $T$ ) towards the pulley and kinetic friction ( $f_k$ ) opposing motion.
Vertical forces balance: $N = m_1g$ , so $f_k = \mu_k N = \mu_k m_1g$ .
Equation of motion: $T - f_k = m_1a \implies T - \mu_k m_1g = m_1a$ ... (ii).

Solve for Acceleration ( $a$ ):

Adding (i) and (ii): $m_2g - \mu_k m_1g = (m_1 + m_2)a$ .
$a = \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2}$ .

Solve for Tension ( $T$ ):

Substitute $a$ into equation (ii): $T = m_1(a + \mu_k g)$ .
$T = m_1 \left[ \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2} + \mu_k g \right]$ .
$T = \frac{m_1 g}{m_1 + m_2} [ m_2 - \mu_k m_1 + \mu_k(m_1 + m_2) ]$ .
$T = \frac{m_1 g}{m_1 + m_2} [ m_2 - \mu_k m_1 + \mu_k m_1 + \mu_k m_2 ]$ .
$T = \frac{m_1 g}{m_1 + m_2} [ m_2 + \mu_k m_2 ] = \frac{m_1 m_2 g (1 + \mu_k)}{m_1 + m_2}$ . (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.9 Friction).

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