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JEE Main · Physics

Laws Of Motion Practice Questions

Practice problems for Laws of Motion in Physics.

147 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q61JEE Main 2022MCQ

A bag is gently dropped on a conveyor belt moving at a speed of $2 {\text{m}} / {\text{s}}$ . The coefficient of friction between the conveyor belt and bag is $0.4$ . Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : $[Take $g=10 {\text{m}} / {\text{s}}^{-2}$ ]$ [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v=2 {\text{m}} / {\text{s}}$ $\mu =0.4$ ${\text{a}}=+(0.4)( {\text{g}})$ $=+4 {\text{m}} / {\text{s}}^2$ ${\text{v}}^{2-} {\text{u}}^2=2$ as $\Rightarrow (4)=2 \times (4)( {\text{s}})$ $s=0.5 {\text{m}}$

Q62JEE Main 2022NAT

Four forces are acting at a point ${P}$ in equilibrium as shown in figure. The ratio of force ${F}_{1}$ to ${F}_{2}$ is $1: x$ where $x=$ [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$F_{1}=+2 \times \;\frac{1}{\sqrt{2}}-\;\frac{1}{\sqrt{2}}=\;\frac{1}{\sqrt{2}}$ $F_{2}=2 \times \;\frac{1}{\sqrt{2}}+\;\frac{1}{\sqrt{2}}=\;\frac{3}{\sqrt{2}}$ $\;\frac{F_{1}}{F_{2}}=\;\frac{1}{3}=\;\frac{1}{x} \Rightarrow x=3$

Explanation Figure 2

Q63JEE Main 2022MCQ

Three masses $M=100 {\text{kg}}, {\text{m}}_1=10 {\text{kg}}$ and ${\text{m}}_2=20 {\text{kg}}$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force ${\text{F}}$ is applied on the system so that the mass ${\text{m}}_2$ moves upward with an acceleration of $2 {\text{ms}}^{-2}$ . The value of ${\text{F}}$ is : $(.$ Take $. {\text{g}}=10 {\text{ms}}^{-2})$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

In frame of block of mass M moving with acceleration a $m_1 a-T=2 m_1 \Rightarrow 10 a-T=20$ $T-m_2 g=m_2 2 \Rightarrow T-200=40 \Rightarrow T=240 . . . . . .$ (ii) $\Rightarrow$ From equation 1 and $210 {\text{a}}=260$ or ${\text{a}}=26 {\text{m}} / {\text{s}}^2$ for block $F(M+m_2) a=120 \times 26$ $=3120 N$

Explanation Figure 2

Explanation Figure 3

Q64JEE Main 2022MCQ

A block of mass $40 {\text{kg}}$ slides over a surface, when a mass of $4 {\text{kg}}$ is suspended through an inextensible massless string passing over frictionless pulley as shown below.The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given ${\text{g}}=10 {\text{ms}}^{-2}$ .) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f_{r_{\max }}=\mu N$ $=0.02 \times 400$ $=8 {\text{N}}$ Let the acceleration is a as shown then. $40-T=4 a$ $T-8=40 a$ $\Rightarrow a=\;\frac{32}{44}=\;\frac{8}{11} {\text{m}} / {\ s}^2$

Explanation Figure 2

Q65JEE Main 2022NAT

A force on an object of mass $100 {\text{g}}$ is $(10 \hat{i}+5 \hat{j}) {\text{N}}$ . The position of that object at ${\text{t}}=2 {\text{s}}$ is $(a \hat{i}+b \hat{j}) {\text{m}}$ after starting from rest. The value of $\;\frac{a}{b}$ will be_____ [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{F}=m \vec{a}$ $\Rightarrow \vec{a}=100 \hat{i}+50 \hat{j}$ So, $\vec{S}=\;\frac{1}{2} \vec{a} t^2$ $\;\frac{1}{2}(100 \hat{i}+50 \hat{j}) 2^2$ $=200 \hat{i}+100 \hat{j} {\text{m}}$ so a $=200 {\text{m}}$ and ${\text{b}}=100 {\text{m}}$ so $\;\frac{a}{b}=2$

Q66JEE Main 2022NAT

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane $A B$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10 {\text{m}}$ . After reaching the Point ${\text{B}}$ the block slides down on inclined plane ${\text{BC}}$ . Time it takes to reach to the point $C$ from point $A$ is $t(\sqrt{2}+1)$ s. The value of $t$ is _______. (use ${\text{g}}=10 {\text{m}} / {\text{s}}^2$ ) [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From E.C. $=\;\frac{1}{2} {\ mv}_0^2= {\text{mgh}}$ ${\text{v}}_0=10 \sqrt{2}$ For ${\text{A}} \to {\text{B}}$ at ${\text{B}}, \;\; {\text{v}}=0$ ${\text{a}}=- {\text{g}} \ sin 45^{\circ}=\;\frac{-10}{\sqrt{2}}$ ${\text{v}}= {\text{u}}+ {\text{at}}_1 \Rightarrow 0=10 \sqrt{2}-\;\frac{10}{\sqrt{2}} {\text{t}}_1 \Rightarrow {\text{t}}_1=2 {\ sec}$ For $B \to C$ ${\ s}=u {\text{t}}_2+\;\frac{1}{2} {\text{at}}_2^2$ $\;\frac{10}{\ sin 30^{\circ}}=\;\frac{1}{2}(10 \ sin 30^{\circ}) {\text{t}}_2^2$ $t_2=2 \sqrt{2}$ So total time ${\text{T}}= {\text{t}}_1+ {\text{t}}_2$ ${\text{T}}= {\text{t}}_1+ {\text{t}}_2$ $=2 \sqrt{2}+2$ $=2(\sqrt{2}+1) {\ sec}$

Q67JEE Main 2022MCQ

Two bodies of masses $m_1=5 {\text{kg}}$ and $m_2=3 {\text{kg}}$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass ${\text{m}}_1$ will be : $[.$ Take $. {\text{g}}=10 {\text{ms}}^{-2}]$ [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Force on equilibrium, $m_2 g=m_1 g sin \;\theta$ Normal force on ${\text{m}}_1, N=m_1 g \cos \theta$ $\Rightarrow \;\frac{N}{m_2 g}=\cot \theta$ $\Rightarrow N=3 \times 10 \times \cot \theta =3 \times 10 \times \;\frac{4}{3} \;\; ( \because sin \;\theta =\;\frac{3}{5})$ $\Rightarrow N=40 {\text{N}}$

Q68JEE Main 2022MCQ

For a free body diagram shown in the figure, the four forces are applied in the ' $x$ ' and ' $y$ ' directions. What additional force must be applied and at what angle with positive $x$ -axis so that the net acceleration of body is zero? [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Resultant of already applied forces $=-\hat{i}-\hat{j}$ $\Rightarrow$ Force required to balance $=\hat{i}+\hat{j}$ $\Rightarrow$ Force required $=\sqrt{2} {N}$ in magnitude at angle $45^{\circ}$ with +ve $x$ -axis

Q69JEE Main 2022MCQ

A block of mass $10 {\text{kg}}$ starts sliding on a surface with an initial velocity of $9.8 {\text{ms}}^{-1}$ . The coefficient of friction between the surface and block is $0.5$ . The distance covered by the block before coming to rest is:$[use ${\text{g}}=9.8 {\text{ms}}^{-2}$ ]$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$S=\;\frac{u^2}{2 a}=\;\frac{u^2}{2(\mu g)}$ $=\;\frac{(9.8)^2}{2 \times 0.5 \times (9.8)}$ $=\;\frac{9.8}{1}$ $=9.8 {\text{m}}$

Q70JEE Main 2022MCQ

A block of mass $M$ slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$ . The magnitude of the contact force will be : [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{N}}= {\text{Mg}} \cos \theta$ ${\text{f}}= {\text{Mg}} \ sin \theta$ ${\text{R}}=\sqrt{ {\text{N}}^{2+} {\text{f}}^2}$ ${\text{R}}= {\text{Mg}}$

Explanation Figure 1

Q71JEE Main 2022NAT

A system to 10 balls each of mass $2 {\text{kg}}$ are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the $7^{\;\text{th }\;}$ and $8^{\;\text{th }\;}$ ball is ____ ${\text{N}}$ when $6^{\;\text{th }\;}$ ball just leaves the table. [26-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

At given instant $a_{\;\text{sys }\;}=\;\frac{6 m \times g}{10 m}=\;\frac{6 g}{10}$ $\therefore T_{78}=(3 m) \times a_{\;\text{sys }\;}$ $=(3 m) \times (\;\frac{6 g}{10})$ $=\;\frac{3 \times 2 \times 6 \times 10}{10}=36 {\text{N}}$

Q72JEE Main 2022MCQ

In the arrangement shown in figure $a_1, a_2, a_3$ and $a_4$ are the accelerations of masses $m_1, m_2, m_3$ and ${\text{m}}_4$ respectively. Which of the following relation is true for this arrangement? [26-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From virtual work done method, $4 T \times a_1+2 T \times a_2+T \times a_3+T \times a_4=0$ $\Rightarrow 4 a_1+2 a_2+a_3+a_4=0$

Q73JEE Main 2022NAT

A hanging mass ${\text{M}}$ is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by $2 {\text{Mg}}$ force. $\ln$ this situation, tension in the string is $\;\frac{x}{5} {\text{Mg}}$ for $x=$ ____. Neglect mass of the string and friction of the block (bigger mass) with ice slab.(Given ${\text{g}}=$ acceleration due to gravity) [28-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$a=\;\frac{M g}{4 M+M}=\;\frac{g}{5}$ (in upward direction) $T=M (g+\;\frac{g}{5}) =\;\frac{6 M g}{5}$ $\Rightarrow x=6$

Q74JEE Main 2022NAT

A mass of $10 {\text{kg}}$ is suspended vertically by a rope of length $5 {\text{m}}$ from the roof. A force of $30 {\text{N}}$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta =\tan ^{-1}(x \times 10^{-1})$ . The value of $x$ is____(Given, $g=10 {\text{m}} / {\text{s}}^2$ ) [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T \cos \theta =m g$ $T \cos \theta =100 N$ $T sin \;\theta =30$ $\Rightarrow \;\frac{T sin \;\theta }{T \cos \theta }=\;\frac{30}{100}$ $\Rightarrow \tan \theta =\;\frac{3}{10}$ $\therefore x=3$

Explanation Figure 1

Q75JEE Main 2022MCQ

An object of mass $5 {\text{kg}}$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 {\text{N}}$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : $[Use ${\text{g}}=10 {\text{ms}}^{-2}$ ]$. [24-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let time taken to ascent is $t_1$ and that to descent is $t_2$ . Height will be same so $H=\;\frac{1}{2} \times 12 t_1^2=\;\frac{1}{2} \times 8 t_2^2$ $\Rightarrow \;\frac{t_1}{t_1}=\;\frac{\sqrt{2}}{\sqrt{3}}$

Q76JEE Main 2022MCQ

A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Using Newton's second law $m g-\;\frac{m g}{4}=m a$ $\Rightarrow a=\;\frac{3 g}{4}$

Explanation Figure 1

Q77JEE Main 2022NAT

A block of mass $200 {\text{g}}$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force ${\text{F}}=$ $\sqrt{x} {\text{N}}$ as shown in figure.The value of $x=____________$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$F \times \;\frac{1}{2}=0.2 \times 10 \times \;\frac{\sqrt{3}}{2}$ $\Rightarrow F=2 \sqrt{3}$ $\Rightarrow F=\sqrt{12} {\text{N}}$ $\therefore x=12$

Explanation Figure 2

Q78JEE Main 2022MCQ

A monkey of mass $50 {\text{kg}}$ climbs on a rope which can withstand the tension (T) of $350 {\text{N}}$ . If monkey initially climbs down with an acceleration of $4 {\text{m}} / {\text{s}}^2$ and then climbs up with an acceleration of $5 {\text{m}} / {\text{s}}^2$ . Choose the correct option $(g=10 {\text{m}} / {\text{s}}^2)$ . [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$T_{\;\text{down }\;}=50 \times (10-4)$ $=50 \times 6$ $=300 {\text{N}}$ ${\text{T}}_{\;\text{up }\;}=50 \times (10+5)$ $=50 \times 15$ $=750 {\text{N}}$ $\Rightarrow$ Rope will break while climbing up.

Q79JEE Main 2022MCQ

A block of mass $2 {\text{kg}}$ moving on a horizontal surface with speed of $4 {\text{ms}}^{-1}$ enters a rough surface ranging from $x=0.5 {\text{m}}$ to $x=1.5 {\text{m}}$ . The retarding force in this range of rough surface is related to distance by $F=-k x$ where $k=12 {\text{Nm}}$ . The speed of the block as it just crosses the rough surface will be : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$F=-12 x$ $m v \;\frac{d v}{d x}=-12 x$ $∫_{4}^{v} v d v=-6 ∫_{0.5}^{1.5} x \text{dx}(m=2 {\text{kg}})$ $\;\frac{v^{2-}16}{2}=-6 [\;\frac{1.5^{2-}0.5^2}{2}]$ $\;\frac{v^{2-}16}{2}=-6$ $v=2 {\text{m}} / {\ sec}$

Q80JEE Main 2021MCQ

A steel block of $10\text{ kg}$ rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2\text{ m/s}^2$ . The normal reaction $R'$ by the floor, if the mass of the iron cylinders are equal and of $20\text{ kg}$ each, is ______ N. $[Take $g = 10\text{ m/s}^2$ and $\mu_s = 0.2$ ]$

🚀 Solve in Practice Mode

📖 Explanation

To find the normal reaction ( $N$ or $R'$ ), we analyze the system (block + 3 cylinders) as a single mass moving downwards.

1. Total Mass of the System ( $M$ ):

Mass of steel block $= 10\text{ kg}$
Mass of 3 cylinders $= 3 \times 20 = 60\text{ kg}$
Total mass $M = 10 + 60 = 70\text{ kg}$

2. Force Equation (Vertical Direction):
The system is accelerating downwards, so the net force is $Mg - N$ .
$Mg - N = Ma$
$N = M(g - a)$

3. Substitution and Calculation:

$M = 70\text{ kg}$
$g = 10\text{ m/s}^2$
$a = 0.2\text{ m/s}^2$

$N = 70 \times (10 - 0.2)$
$N = 70 \times 9.8$
$N = 686\text{ N}$

Note: Since the motion is vertical and there is no horizontal force pushing the block against a wall, the coefficient of static friction ( $\mu_s$ ) does not affect the normal reaction.

Correct Option: (B)

Explanation Figure 2

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