Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block of mass $m$ is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$ . The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies:

$\alpha > \frac{mg}{\mu}$

$\alpha > \frac{g}{\mu m}$

$\alpha \ge \frac{g}{\mu}$

$\alpha < \frac{g}{\mu}$

Step-by-Step Solution

Identify Forces:

Vertical Direction: The block has a tendency to fall downwards due to gravity ( $mg$ ). To prevent this, the static friction force ( $f_s$ ) must act upwards.
Horizontal Direction: The cart accelerates with $\alpha$ . The block moves with the cart, so a normal reaction force ( $N$ ) from the cart acts on the block to providing the necessary acceleration. $N = m\alpha$ [NCERT Class 11, Physics Part I, Sec 5.5].

Condition for Equilibrium (No slipping): For the block not to fall, the upward friction must balance the downward weight: $f_s = mg$ .
Limiting Friction: The static friction is self-adjusting but has a maximum limit given by $f_s \le \mu_s N$ (where $\mu$ is the coefficient of static friction) [NCERT Class 11, Physics Part I, Sec 5.9, Eq 4.14].
Derivation: Substitute the values into the inequality: $mg \le \mu N$ $mg \le \mu (m\alpha)$ Cancel mass $m$ from both sides (since $m > 0$ ): $g \le \mu \alpha$ $\alpha \ge \frac{g}{\mu}$

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