Laws of Motion — NEET UG practice

1. When two forces act perpendicular to each other on a body, the net (resultant) force is found using the Pythagorean rule, since perpendicular vectors combine like the two legs of a right triangle.

Given forces $F_1 = 8 \text{ N}$ and $F_2 = 6 \text{ N}$,

$$F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10 \text{ N}$$

2. By Newton's second law, acceleration equals the net force divided by mass.

Therefore, with $m = 5 \text{ kg}$,

$$a = \frac{F_{net}}{m} = \frac{10}{5} = 2 \text{ m/s}^2$$

3. To find the direction of this resultant relative to the $8$ N force, we use the fact that the angle a resultant makes with one component follows $\tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}}$, where the $6$ N force acts perpendicular to the $8$ N force.

So,

$$\tan\theta = \frac{6}{8} = \frac{3}{4}$$

4. Taking the inverse tangent gives the angle measured from the $8$ N force.

Hence,

$$\theta = \tan^{-1}\left(\frac{3}{4}\right) \text{ from the } 8 \text{ N force}$$

Hence, the answer is Option D.

1. When a road is banked at the ideal angle for a given speed, the car doesn't need friction from the tyres to provide the centripetal force — this is exactly the condition for minimum wear and tear.

Given that friction plays no role, only gravity and the normal reaction supply the centripetal force, so the standard banking formula applies here.

$$\tan\theta = \frac{v^2}{rg}$$

2. Substitute the given values $v = 10 \text{ ms}^{-1}$, $r = 50 \text{ m}$, and $g = 10 \text{ ms}^{-2}$.

Therefore,

$$\tan\theta = \frac{10^2}{50 \times 10} = \frac{100}{500} = \frac{1}{5}$$

3. Taking the inverse tangent on both sides gives the banking angle.

Hence,

$$\theta = \tan^{-1}\left(\frac{1}{5}\right)$$

Hence, the answer is Option B.

1. For the box to remain stationary relative to the trolley, the static friction force between the box and the trolley floor must be the only horizontal force accelerating the box.

The maximum static friction available is $\mu N$, and since the box only experiences gravity and the normal reaction vertically, $N = mg$.

2. The box will start slipping the moment the required force exceeds this maximum friction, so the maximum possible acceleration occurs when friction is at its limiting value.

Therefore, applying Newton's second law to the box,

$$\mu mg = ma_{max}$$

3. Since mass $m$ appears on both sides, it cancels out, leaving acceleration dependent only on $\mu$ and $g$.

Hence,

$$a_{max} = \mu g$$

4. Substituting $\mu = 0.12$ and $g = 10 \text{ m/s}^2$.

So,

$$a_{max} = 0.12 \times 10 = 1.2 \text{ m/s}^2$$

Hence, the answer is Option D.