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 $a$ of the cart that will prevent the block from falling satisfies:

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

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

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

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

Step-by-Step Solution

Forces Analysis: Consider the block of mass $m$ against the vertical surface of the accelerating cart.

Horizontal Direction: The cart accelerates with acceleration $a$ . The normal reaction $N$ provided by the vertical surface of the cart on the block causes this acceleration. According to Newton's Second Law: $N = ma$ .
Vertical Direction: The weight $mg$ acts downwards. To prevent the block from falling, the static friction force $f_s$ acting upwards must balance the weight. Thus, $f_s = mg$ .

Friction Condition: For the block not to slip, the required static friction must be less than or equal to the limiting value ( $f_s \le f_{s,max}$ ). The limiting friction is given by $f_{s,max} = \mu N$ [NCERT Class 11, Physics Part I, Section 5.9 Friction].
Derivation: $f_s \le \mu N$ Substitute $f_s = mg$ and $N = ma$ : $mg \le \mu (ma)$ Cancel the mass $m$ (since $m > 0$ ): $g \le \mu a$ $a \ge \frac{g}{\mu}$ Thus, the minimum acceleration required is $g/\mu$ .

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