Maximum friction force acting on the object will be uN= 0.3 * 30 = 9 N

But applied force is 5N so friction force acting on the object will be 5N

Given

$m_1 = 5$

kg,

$m_2 = 3$

kg, and

$g = 10$

m/s²:

• Tension:
  $T = m_2 g = 3 \times 10 = 30$ 

  N

• Friction force:
  $f = \mu N = \mu \times 50$ 
• At equilibrium:
  $\mu \times 50 = 30$ 

Solving,

$\mu = \frac{30}{50} = 0.6$

.

The condition to prevent the block from falling is that the friction force must be at least equal to the weight of the block:

$$f_s \geq mg$$

Since static friction is given by $f_s = \mu N$ and the normal force is due to pseudo force $N = ma$, we get:

$$\mu ma \geq mg$$

Dividing both sides by $\mu m$:

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

Thus, the required acceleration of the cart is $a \geq \frac{g}{\mu}$.

The forces acting on the body are: Normal reaction (N$=m\mathrm{g)}$(upward) and Friction force f$\le \mu mg$(opposing applied force)

The resultant force is:

$$F = \sqrt{N^2 + f^2} = \sqrt{mg^2 + (\mu mg)^2} = mg \sqrt{1 + \mu^2}$$

$F \leq mg(1 + \mu^2)^{1/2}$

The stopping distance is given by \(s = \frac{v^2}{2a}\), where the frictional retardation is \(a = \mu g\). Substituting the relation for momentum \(v = \frac{P}{m}\) into the formula yields \(s = \frac{P^2}{2\mu m^2g}\).

The maximum limiting frictional force is \(f_{max} = \mu m g = 0.2 \times 20 \times 10 = 40\text{ N}\) (using \(g = 10\text{ m/s}^2\)). Since the applied force of \(40\text{ N}\) is equal to the limiting friction, the net horizontal force on the block is zero, resulting in zero acceleration.

Limiting friction is the maximum value of static friction (A \(\rightarrow\) Q). Static friction is a self-adjusting force (B \(\rightarrow\) R). Kinetic friction has a constant magnitude of \(\mu_k N\) (C \(\rightarrow\) P).

Assertion (A) is true: To walk, a person pushes backward on the ground. By Newton's third law, the ground exerts a forward static friction force on the person's feet. This forward friction, combined with the normal force, creates a net force from the surface in the direction of motion, propelling the person forward.nReason (R) is false: For walking, the friction force exerted by the road on the person's feet is *in the direction* of the person's motion (forward friction), enabling propulsion. If friction were against the person's motion, walking would be impossible. Therefore, (A) is true but (R) is false.

Assertion (A) is true: In an emergency, applying brakes in a straight line is generally a safer and more controlled maneuver to reduce speed and avoid collision, as sudden turns at high speed can lead to loss of control or skidding.

Reason (R) is true: Both stopping the car (through braking) and taking a turn (requiring centripetal force) on a horizontal road fundamentally rely on the friction force between the tires and the road. Reason (R) is true, but it does not explain *why* braking is preferred over turning; it merely states that friction is involved in both actions.

Therefore, (A) and (R) are true, but (R) is not the correct explanation of (A).

Assertion (A) is true due to low friction on ice. Reason (R) is also true as small steps are needed to minimize horizontal forces and prevent slipping. (R) explains why extra effort and carefulness are required, making the activity tiring.