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:

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

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