NEET Laws of Motion MCQs & PYQs

Q: If block is at verge of motion, then select the correct option (where μ is the coefficient of friction)

If the angle θ\theta is with the vertical, then we need to adjust the force components accordingly: Normal Force: N=mg−Fcos⁡θN = mg - F \cos \theta Friction Force: fs=μN=μ(mg−Fcos⁡θ)f_s = \mu N = \mu (mg - F \cos \theta) Horizontal Component of Force: Fx=Fsin⁡θF_x = F \sin \theta At the verge of motion: Fsin⁡θ=μ(mg−Fcos⁡θ)F \sin \theta = \mu (mg - F \cos \theta) Solving for FF: F=μmgsin⁡θ+μcos⁡θF = \frac{\mu mg}{\sin \theta + \mu \cos \theta}

Q: The minimum value of coefficient of friction (μ) such that block of mass ‘5 kg’ remains at rest is

Given m1=5m_1 = 5 kg, m2=3m_2 = 3 kg, and g=10g = 10 m/s²: Tension: T=m2g=3×10=30T = m_2 g = 3 \times 10 = 30 N Friction force: f=μN=μ×50f = \mu N = \mu \times 50 At equilibrium: μ×50=30\mu \times 50 = 30 Solving, μ=3050=0.6\mu = \frac{30}{50} = 0.6 .

Q: The position-time graph of a particle of mass 2 kg moving along x-axis is as shown in the figure. The magnitude of impulse on the particle at t = 2 s is

Solution: Impulse is given by the change in momentum: J=Δp=mΔvJ = \Delta p = m \Delta v From the graph, we find velocity before and after t=2st = 2s: Before t=2st = 2s: vinitial=x2−x1t2−t1=0−202−0=−10 m/sv_{\text{initial}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{0 - 20}{2 - 0} = -10 \text{ m/s} After t=2st = 2s: vfinal=x3−x2t3−t2=20−04−2=10 m/sv_{\text{final}} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{20 - 0}{4 - 2} = 10 \text{ m/s} Now, impulse: J=m(vfinal−vinitial) J = m (v_{\text{final}} - v_{\text{initial}}) J=2×(10−(−10))=2×20=40 NsJ = 2 \times (10 - (-10)) = 2 \times 20 = 40 \text{ Ns} Final Answer: 40 N-s\mathbf{40 \text{ Ns}}

Q: A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of a. The pendulum will tilt

When the truck accelerates to the right with aa, a pseudo force mama acts to the left on the bob in the truck's frame. The bob reaches equilibrium where the tension components balance forces: tan⁡θ=Pseudo forceWeight=mamg=ag\tan \theta = \frac{\text{Pseudo force}}{\text{Weight}} = \frac{ma}{mg} = \frac{a}{g} θ=tan⁡−1(ag)\theta = \tan^{-1} \left(\frac{a}{g} \right) Thus, the pendulum tilts to the left at an angle θ=tan⁡−1(a/g)\theta = \tan^{-1}(a/g) with the vertical.

Q: A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where F is:

The forces acting on the body are: Normal reaction (N=mg) (upward) and Friction force f≤μmg (opposing applied force) The resultant force is: F=N2+f2=mg2+(μmg)2=mg1+μ2F = \sqrt{N^2 + f^2} = \sqrt{mg^2 + (\mu mg)^2} = mg \sqrt{1 + \mu^2} F \leq mg(1 + \mu^2)^{1/2}

Practice NEET Laws of Motion Questions

Question 61:

If block is at verge of motion, then select the correct option (where μ is the coefficient of friction)

neet friction based queestion

If the angle $\theta$ is with the vertical, then we need to adjust the force components accordingly:

Normal Force:
$N = mg - F \cos \theta$
Friction Force:
$f_s = \mu N = \mu (mg - F \cos \theta)$
Horizontal Component of Force:
$F_x = F \sin \theta$

At the verge of motion:

$F \sin \theta = \mu (mg - F \cos \theta)$

Solving for $F$ :

$F = \frac{\mu mg}{\sin \theta + \mu \cos \theta}$

💬 Discuss this Question

Question 62:

easy

In translatory equilibrium

In translational equilibrium the net force acting on the object is zero. so the object moves with constant velocity.

💬 Discuss this Question

Question 63:

easy

The minimum value of coefficient of friction (μ) such that block of mass ‘5 kg’ remains at rest is

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,

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

💬 Discuss this Question

Question 64:

If the coefficient of friction between the block of mass 5 kg and wall is 0.5, then minimum force F required to hold the block with the wall is
(g = 10 m/s²)

\[
\text{Given:} \quad m = 5 \text{ kg}, \quad g = 10 \text{ m/s}^2, \quad \mu = 0.5
\]

\text{Normal force:}
\[
N = F
\]

\text{Friction force:}
\[
f = \mu N = \mu F
\]

\text{For equilibrium:}
\[
\mu F = mg
\]

\text{Substituting values:}
\[
0.5 F = 5 \times 10
\]

\[
0.5 F = 50
\]

\[
F = \frac{50}{0.5} = 100 \text{ N}
\]

\textbf{Final Answer:}
\[
\mathbf{F = 100 N}
\]

💬 Discuss this Question

Question 65:

easy

Sand is poured on a conveyor belt at the rate of 2 kg/s. If belt is moving horizontally with velocity 4 m/s, then additional force required by engine to keep the belt moving with same constant velocity

The force required to keep the conveyor belt moving at a constant velocity is given by the rate of change of momentum.

Momentum of sand per second:

$\text{Force} = \frac{\text{dP}}{\text{dt}} = \frac{\text{dm}}{\text{dt}} \times v$

Given $\frac{dm}{dt} = 2$ kg/s and $v = 4$ m/s,

$F = 2 \times 4 = 8 \text{ N}$

Thus, the additional force required is 8 N.

💬 Discuss this Question

Question 66:

The position-time graph of a particle of mass 2 kg moving along x-axis is as shown in the figure. The magnitude of impulse on the particle at t = 2 s is

Solution:

Impulse is given by the change in momentum:

$J = \Delta p = m \Delta v$

From the graph, we find velocity before and after $t = 2s$ :

Before $t = 2s$ :
$v_{\text{initial}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{0 - 20}{2 - 0} = -10 \text{ m/s}$
After $t = 2s$ :
$v_{\text{final}} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{20 - 0}{4 - 2} = 10 \text{ m/s}$

Now, impulse:

$J = m (v_{final} - v_{initial})$

$J = m (v_{\text{final}} - v_{\text{initial}})$ $J = 2 \times (10 - (-10)) = 2 \times 20 = 40 \text{ Ns}$

Final Answer:

$\mathbf{40 \text{ Ns}}$

💬 Discuss this Question

Question 67:

easy

The force F needed to keep the block at equilibrium in given figure is (pulley and string are massless)

neet constrained motion questions

For Equilibrium,

2F = Mg ⇒ F = Mg/2

💬 Discuss this Question

Question 68:

easy

A block of mass m is in contact with the cart. The coefficient of static friction between the block and the cart is ü. The acceleration a of the cart that prevent the block from falling will be

neet pseudo force questions

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

💬 Discuss this Question

Question 69:

easy

A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of a. The pendulum will tilt

When the truck accelerates to the right with $a$ , a pseudo force $ma$ acts to the left on the bob in the truck's frame. The bob reaches equilibrium where the tension components balance forces:

$\tan \theta = \frac{\text{Pseudo force}}{\text{Weight}} = \frac{ma}{mg} = \frac{a}{g}$ $\theta = \tan^{-1} \left(\frac{a}{g} \right)$

Thus, the pendulum tilts to the left at an angle $\theta = \tan^{-1}(a/g)$ with the vertical.

💬 Discuss this Question

Question 70:

easy

A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where F is:

The forces acting on the body are: Normal reaction (N $= m g)$ (upward) and Friction force f $\leq μ m g$ (opposing applied force)

The resultant force is:

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

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

💬 Discuss this Question