📘 Laws of Motion — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Laws of Motion for JEE Main with previous year questions on Newton's laws, friction, and circular motion. Each solution is explained in a simple, step-by-step manner to boost your score.
Laws of Motion — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 1)
A cubic block of mass \(m\) is sliding down on an inclined plane at \(60^\circ\) with an acceleration of \(g/2\), the value of coefficient of kinetic friction is :
(1) \(\sqrt{3} - 1\) (2) \(\frac{\sqrt{3}}{2}\)
(3) \(\frac{\sqrt{2}}{3}\) (4) \(1 - \frac{\sqrt{3}}{2}\)
(1) \(\sqrt{3} - 1\) (2) \(\frac{\sqrt{3}}{2}\)
(3) \(\frac{\sqrt{2}}{3}\) (4) \(1 - \frac{\sqrt{3}}{2}\)
✅ Answer & Explanation:
Correct Option: (1)
For a block sliding down an incline:
\(mg \sin\theta - \mu_k mg \cos\theta = ma\)
Substituting \(\theta = 60^\circ\) and \(a = g/2\):
\(g \sin 60^\circ - \mu_k g \cos 60^\circ = g/2\)
\(\frac{\sqrt{3}}{2} - \mu_k \frac{1}{2} = \frac{1}{2} \implies \mu_k = \sqrt{3} - 1\).
For a block sliding down an incline:
\(mg \sin\theta - \mu_k mg \cos\theta = ma\)
Substituting \(\theta = 60^\circ\) and \(a = g/2\):
\(g \sin 60^\circ - \mu_k g \cos 60^\circ = g/2\)
\(\frac{\sqrt{3}}{2} - \mu_k \frac{1}{2} = \frac{1}{2} \implies \mu_k = \sqrt{3} - 1\).
Q2. JEE Main 2025 (4 April Shift 1)
A body of mass \(m\) is suspended by two strings making angles \(\theta_1\) and \(\theta_2\) with the horizontal ceiling with tensions \(T_1\) and \(T_2\) simultaneously. \(T_1\) and \(T_2\) are related by \(T_1 = \sqrt{3}T_2\). The angles \(\theta_1\) and \(\theta_2\) are :
(1) \(\theta_1 = 30^\circ, \theta_2 = 60^\circ\) with \(T_2 = \frac{3mg}{4}\)
(2) \(\theta_1 = 60^\circ, \theta_2 = 30^\circ\) with \(T_2 = \frac{mg}{2}\)
(3) \(\theta_1 = 45^\circ, \theta_2 = 45^\circ\) with \(T_2 = \frac{3mg}{4}\)
(4) \(\theta_1 = 30^\circ, \theta_2 = 60^\circ\) with \(T_2 = \frac{4mg}{5}\)
(2) \(\theta_1 = 60^\circ, \theta_2 = 30^\circ\) with \(T_2 = \frac{mg}{2}\)
(3) \(\theta_1 = 45^\circ, \theta_2 = 45^\circ\) with \(T_2 = \frac{3mg}{4}\)
(4) \(\theta_1 = 30^\circ, \theta_2 = 60^\circ\) with \(T_2 = \frac{4mg}{5}\)
✅ Answer & Explanation:
Correct Option: (2)
From horizontal equilibrium: \(T_1 \cos\theta_1 = T_2 \cos\theta_2\).
Given \(T_1 = \sqrt{3}T_2\), we get \(\sqrt{3} \cos\theta_1 = \cos\theta_2\).
Checking Option (2): \(\theta_1 = 60^\circ, \theta_2 = 30^\circ \implies \sqrt{3}(1/2) = \sqrt{3}/2\). (Matches)
From vertical equilibrium: \(T_1 \sin 60^\circ + T_2 \sin 30^\circ = mg\)
\(\sqrt{3}T_2 (\sqrt{3}/2) + T_2 (1/2) = mg \implies 2T_2 = mg \implies T_2 = mg/2\).
From horizontal equilibrium: \(T_1 \cos\theta_1 = T_2 \cos\theta_2\).
Given \(T_1 = \sqrt{3}T_2\), we get \(\sqrt{3} \cos\theta_1 = \cos\theta_2\).
Checking Option (2): \(\theta_1 = 60^\circ, \theta_2 = 30^\circ \implies \sqrt{3}(1/2) = \sqrt{3}/2\). (Matches)
From vertical equilibrium: \(T_1 \sin 60^\circ + T_2 \sin 30^\circ = mg\)
\(\sqrt{3}T_2 (\sqrt{3}/2) + T_2 (1/2) = mg \implies 2T_2 = mg \implies T_2 = mg/2\).
Q3. JEE Main 2024 (31 Jan Shift 2)
A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If \(\vec{F}_1\) is the force required to just move the block up the inclined plane and \(\vec{F}_2\) is the force required to just prevent the block from sliding down, then the value of \(|\vec{F}_1 - \vec{F}_2|\) is: [Use \(g = 10 \, \text{m/s}^2\)]
(1) \(25\sqrt{3} \, \text{N}\) (2) \(5\sqrt{3} \, \text{N}\)
(3) \(\frac{5\sqrt{3}}{2} \, \text{N}\) (4) \(10 \, \text{N}\)
(3) \(\frac{5\sqrt{3}}{2} \, \text{N}\) (4) \(10 \, \text{N}\)
✅ Answer & Explanation:
Correct Option: (2)
- \(F_{gravity} = mg \sin 30^\circ = 5 \times 10 \times 0.5 = 25 \, \text{N}\).
- \(f_{friction} = \mu mg \cos 30^\circ = 0.1 \times 5 \times 10 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \, \text{N}\).
- \(F_1 = F_{gravity} + f_{friction} = 25 + 2.5\sqrt{3}\).
- \(F_2 = F_{gravity} - f_{friction} = 25 - 2.5\sqrt{3}\).
- \(|F_1 - F_2| = 2 \times f_{friction} = 5\sqrt{3} \, \text{N}\).
- \(F_{gravity} = mg \sin 30^\circ = 5 \times 10 \times 0.5 = 25 \, \text{N}\).
- \(f_{friction} = \mu mg \cos 30^\circ = 0.1 \times 5 \times 10 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \, \text{N}\).
- \(F_1 = F_{gravity} + f_{friction} = 25 + 2.5\sqrt{3}\).
- \(F_2 = F_{gravity} - f_{friction} = 25 - 2.5\sqrt{3}\).
- \(|F_1 - F_2| = 2 \times f_{friction} = 5\sqrt{3} \, \text{N}\).
Q4. JEE Main 2023 (01 Feb Shift 1)
A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given, \(g = 10 \, \text{m/s}^2\)) :
(1) 0.60 (2) 0.75
(3) 0.50 (4) 0.25
(1) 0.60 (2) 0.75
(3) 0.50 (4) 0.25
✅ Answer & Explanation:
Correct Option: (3)
- \(s = \frac{1}{2}at^2 \implies 50 = \frac{1}{2}a(100) \implies a = 1 \, \text{m/s}^2\).
- \(F - \mu_k mg = ma \implies 30 - \mu_k(50) = 5(1)\).
- \(50\mu_k = 25 \implies \mu_k = 0.50\).
- \(s = \frac{1}{2}at^2 \implies 50 = \frac{1}{2}a(100) \implies a = 1 \, \text{m/s}^2\).
- \(F - \mu_k mg = ma \implies 30 - \mu_k(50) = 5(1)\).
- \(50\mu_k = 25 \implies \mu_k = 0.50\).
Q5. JEE Main 2022 (25 Jul Shift 1)
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 \(x = \) _____.
✅ Answer & Explanation:
Correct Answer: 3
Resolving forces along X and Y axes:
- \(\sum F_x = 0 \implies F_1 + 1\cos 45^\circ - 2\cos 45^\circ = 0 \implies F_1 = \frac{1}{\sqrt{2}}\).
- \(\sum F_y = 0 \implies 1\sin 45^\circ + 2\sin 45^\circ - F_2 = 0 \implies F_2 = \frac{3}{\sqrt{2}}\).
Ratio \(F_1 / F_2 = 1/3\). Thus \(x = 3\).
Resolving forces along X and Y axes:
- \(\sum F_x = 0 \implies F_1 + 1\cos 45^\circ - 2\cos 45^\circ = 0 \implies F_1 = \frac{1}{\sqrt{2}}\).
- \(\sum F_y = 0 \implies 1\sin 45^\circ + 2\sin 45^\circ - F_2 = 0 \implies F_2 = \frac{3}{\sqrt{2}}\).
Ratio \(F_1 / F_2 = 1/3\). Thus \(x = 3\).
Q6. JEE Main 2021 (17 Mar Shift 1)
Two blocks (\(m = 0.5 \, \text{kg}\) and \(M = 4.5 \, \text{kg}\)) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is _____ N. (Round off to the Nearest Integer) [Take \(g\) as \(9.8 \, \text{m/s}^2\)]
✅ Answer & Explanation:
Correct Answer: 21
- Max friction between blocks \(f = \mu mg = \frac{3}{7} \times 0.5 \times 9.8 = 2.1 \, \text{N}\).
- Max acceleration of block \(m\): \(a = f/m = 2.1 / 0.5 = 4.2 \, \text{m/s}^2\).
- For both blocks to move together: \(F = (M+m)a = (4.5 + 0.5) \times 4.2 = 21 \, \text{N}\).
- Max friction between blocks \(f = \mu mg = \frac{3}{7} \times 0.5 \times 9.8 = 2.1 \, \text{N}\).
- Max acceleration of block \(m\): \(a = f/m = 2.1 / 0.5 = 4.2 \, \text{m/s}^2\).
- For both blocks to move together: \(F = (M+m)a = (4.5 + 0.5) \times 4.2 = 21 \, \text{N}\).
Q7. JEE Main 2025 (23 Jan Shift 2)
A massless spring gets elongated by amount \(x_1\) under a tension of 5 N. Its elongation is \(x_2\) under the tension of 7 N. For the elongation of \((5x_1 - 2x_2)\), the tension in the spring will be :
(1) 39 N (2) 15 N
(3) 11 N (4) 20 N
(1) 39 N (2) 15 N
(3) 11 N (4) 20 N
✅ Answer & Explanation:
Correct Option: (3)
Using Hooke's Law \(F = kx\):
\(5 = kx_1\) and \(7 = kx_2\).
Required tension \(T = k(5x_1 - 2x_2)\)
\(T = 5(kx_1) - 2(kx_2) = 5(5) - 2(7) = 25 - 14 = 11 \, \text{N}\).
Using Hooke's Law \(F = kx\):
\(5 = kx_1\) and \(7 = kx_2\).
Required tension \(T = k(5x_1 - 2x_2)\)
\(T = 5(kx_1) - 2(kx_2) = 5(5) - 2(7) = 25 - 14 = 11 \, \text{N}\).
Q8. JEE Main 2025 (2 April Shift 2)
A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions \(T_1\) and \(T_2\), respectively, are (in N) :
(1) \(5, 5\sqrt{3}\) (2) \(5\sqrt{3}, 5\)
(3) \(5\sqrt{3}, 5\sqrt{3}\) (4) \(5, 5\)
(3) \(5\sqrt{3}, 5\sqrt{3}\) (4) \(5, 5\)
✅ Answer & Explanation:
Correct Option: (2)
Applying Lami's theorem or resolution of forces:
\(T_1 \cos 60^\circ = T_2 \cos 30^\circ \implies T_1(1/2) = T_2(\sqrt{3}/2) \implies T_1 = \sqrt{3}T_2\).
\(T_1 \sin 60^\circ + T_2 \sin 30^\circ = mg = 10\)
\((\sqrt{3}T_2) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10 \implies 2T_2 = 10 \implies T_2 = 5 \, \text{N}\).
Then \(T_1 = 5\sqrt{3} \, \text{N}\).
Applying Lami's theorem or resolution of forces:
\(T_1 \cos 60^\circ = T_2 \cos 30^\circ \implies T_1(1/2) = T_2(\sqrt{3}/2) \implies T_1 = \sqrt{3}T_2\).
\(T_1 \sin 60^\circ + T_2 \sin 30^\circ = mg = 10\)
\((\sqrt{3}T_2) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10 \implies 2T_2 = 10 \implies T_2 = 5 \, \text{N}\).
Then \(T_1 = 5\sqrt{3} \, \text{N}\).
Q9. JEE Main 2024 (31 Jan Shift 1)
In the given arrangement of a doubly inclined plane two blocks of masses \(M\) and \(m\) are placed. The coefficient of friction between the surface and the blocks is 0.25. The value of \(m\), for which \(M = 10 \, \text{kg}\) will move down with an acceleration of \(2 \, \text{m/s}^2\), is: (take \(g = 10 \, \text{m/s}^2\) and \(\tan 37^\circ = 3/4\))
(1) 9 kg (2) 4.5 kg
(3) 6.5 kg (4) 2.25 kg
(3) 6.5 kg (4) 2.25 kg
✅ Answer & Explanation:
Correct Option: (2)
Writing equations for both blocks and solving for \(m\) yields **4.5 kg**.
Writing equations for both blocks and solving for \(m\) yields **4.5 kg**.
Q10. JEE Main 2024 (06 Apr Shift 1)
A light string passing over a smooth light pulley connects two blocks of masses \(m_1\) and \(m_2\) (where \(m_2 > m_1\)). If the acceleration of the system is \(g/\sqrt{2}\), then the ratio of the masses \(m_1/m_2\) is :
(1) \(\frac{1+\sqrt{5}}{\sqrt{5}-1}\) (2) \(\frac{\sqrt{2}-1}{\sqrt{2}+1}\)
(3) \(\frac{1+\sqrt{5}}{\sqrt{2}-1}\) (4) \(\frac{\sqrt{3}+1}{\sqrt{2}-1}\)
(1) \(\frac{1+\sqrt{5}}{\sqrt{5}-1}\) (2) \(\frac{\sqrt{2}-1}{\sqrt{2}+1}\)
(3) \(\frac{1+\sqrt{5}}{\sqrt{2}-1}\) (4) \(\frac{\sqrt{3}+1}{\sqrt{2}-1}\)
✅ Answer & Explanation:
Correct Option: (2)
Acceleration \(a = \frac{m_2 - m_1}{m_1 + m_2} g\).
Given \(a = g/\sqrt{2} \implies \frac{1}{\sqrt{2}} = \frac{m_2 - m_1}{m_2 + m_1}\).
Using componendo and dividendo:
\(\frac{m_1}{m_2} = \frac{\sqrt{2}-1}{\sqrt{2}+1}\).
Acceleration \(a = \frac{m_2 - m_1}{m_1 + m_2} g\).
Given \(a = g/\sqrt{2} \implies \frac{1}{\sqrt{2}} = \frac{m_2 - m_1}{m_2 + m_1}\).
Using componendo and dividendo:
\(\frac{m_1}{m_2} = \frac{\sqrt{2}-1}{\sqrt{2}+1}\).
Q11. JEE Main 2024 (05 Apr Shift 1)
Three blocks \(M_1, M_2, M_3\) having masses 4 kg, 6 kg and 10 kg respectively are hanging from a smooth pulley using rope 1, 2 and 3 as shown in figure. The tension in the rope 1, \(T_1\) when they are moving upward with acceleration of \(2 \, \text{m/s}^2\) is _____ N (if \(g = 10 \, \text{m/s}^2\)).
✅ Answer & Explanation:
Correct Answer: 240
Tension \(T_1\) supports the total mass of the system.
\(T_1 = (M_1 + M_2 + M_3)(g + a)\)
\(T_1 = (4 + 6 + 10)(10 + 2) = 20 \times 12 = 240 \, \text{N}\).
Tension \(T_1\) supports the total mass of the system.
\(T_1 = (M_1 + M_2 + M_3)(g + a)\)
\(T_1 = (4 + 6 + 10)(10 + 2) = 20 \times 12 = 240 \, \text{N}\).
Q12. JEE Main 2022 (26 Jul Shift 1)
Three masses \(M = 100 \, \text{kg}, m_1 = 10 \, \text{kg}\) and \(m_2 = 20 \, \text{kg}\) 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 \(F\) is applied on the system so that the mass \(m_2\) moves upward with an acceleration of \(2 \, \text{m/s}^2\). The value of \(F\) is (Take \(g = 10 \, \text{m/s}^2\)) :
(1) 3360 N (2) 3380 N
(3) 3120 N (4) 3240 N
(3) 3120 N (4) 3240 N
✅ Answer & Explanation:
Correct Option: (1)
Solving the system using Newton's laws for each block leads to a required force of **3360 N**.
Solving the system using Newton's laws for each block leads to a required force of **3360 N**.
Q13. JEE Main 2025 (24 Jan Shift 2)
A string of length \(L\) is fixed at one end and carries a mass of \(M\) at the other end. The mass makes \((3/\pi)\) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ______ \(ML\).
✅ Answer & Explanation:
Correct Answer: 36
For a conical pendulum: \(T \sin\theta = M \omega^2 R\).
Since \(R = L \sin\theta\), we have \(T \sin\theta = M \omega^2 L \sin\theta \implies T = M \omega^2 L\).
Frequency \(f = 3/\pi \, \text{Hz} \implies \omega = 2\pi f = 6 \, \text{rad/s}\).
\(T = M(6^2)L = 36 ML\).
For a conical pendulum: \(T \sin\theta = M \omega^2 R\).
Since \(R = L \sin\theta\), we have \(T \sin\theta = M \omega^2 L \sin\theta \implies T = M \omega^2 L\).
Frequency \(f = 3/\pi \, \text{Hz} \implies \omega = 2\pi f = 6 \, \text{rad/s}\).
\(T = M(6^2)L = 36 ML\).
Q14. JEE Main 2025 (24 Jan Shift 1)
A car of mass '\(m\)' moves on a banked road having radius '\(r\)' and banking angle \(\theta\). To avoid slipping from banked road, the maximum permissible speed of the car is \(v_0\). The coefficient of friction \(\mu\) between the wheels of the car and the banked road is :
(1) \(\mu = \frac{v_0^2 + rg \tan\theta}{rg + v_0^2 \tan\theta}\) (2) \(\mu = \frac{v_0^2 - rg \tan\theta}{rg - v_0^2 \tan\theta}\)
(3) \(\mu = \frac{v_0^2 - rg \tan\theta}{rg + v_0^2 \tan\theta}\) (4) \(\mu = \frac{v_0^2 + rg \tan\theta}{rg - v_0^2 \tan\theta}\)
(1) \(\mu = \frac{v_0^2 + rg \tan\theta}{rg + v_0^2 \tan\theta}\) (2) \(\mu = \frac{v_0^2 - rg \tan\theta}{rg - v_0^2 \tan\theta}\)
(3) \(\mu = \frac{v_0^2 - rg \tan\theta}{rg + v_0^2 \tan\theta}\) (4) \(\mu = \frac{v_0^2 + rg \tan\theta}{rg - v_0^2 \tan\theta}\)
✅ Answer & Explanation:
Correct Option: (3)
The maximum speed on a banked road with friction is:
\(v_0 = \sqrt{rg \frac{\mu + \tan\theta}{1 - \mu \tan\theta}}\)
\(\frac{v_0^2}{rg} = \frac{\mu + \tan\theta}{1 - \mu \tan\theta} \implies v_0^2 - \mu v_0^2 \tan\theta = \mu rg + rg \tan\theta\)
\(\mu(rg + v_0^2 \tan\theta) = v_0^2 - rg \tan\theta \implies \mu = \frac{v_0^2 - rg \tan\theta}{rg + v_0^2 \tan\theta}\).
The maximum speed on a banked road with friction is:
\(v_0 = \sqrt{rg \frac{\mu + \tan\theta}{1 - \mu \tan\theta}}\)
\(\frac{v_0^2}{rg} = \frac{\mu + \tan\theta}{1 - \mu \tan\theta} \implies v_0^2 - \mu v_0^2 \tan\theta = \mu rg + rg \tan\theta\)
\(\mu(rg + v_0^2 \tan\theta) = v_0^2 - rg \tan\theta \implies \mu = \frac{v_0^2 - rg \tan\theta}{rg + v_0^2 \tan\theta}\).
Q15. JEE Main 2023 (30 Jan Shift 1)
The figure represents the momentum time (\(p-t\)) curve for a particle moving along an axis under the influence of the force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively? If \((t_3 - t_2) < t_1\)
(1) c and a (2) b and c
(3) c and b (4) a and b
(3) c and b (4) a and b
✅ Answer & Explanation:
Correct Option: (3)
Force is the rate of change of momentum: \(F = dp/dt\) (slope of the curve).
The slope is maximum in region **c** and minimum (flattest) in region **b**.
Force is the rate of change of momentum: \(F = dp/dt\) (slope of the curve).
The slope is maximum in region **c** and minimum (flattest) in region **b**.
Q16. JEE Main 2022 (29 Jul Shift 1)
A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass \(m\) moves against the wall with a speed \(v\). Which of the following curve represents the correct relation between the normal reaction on the block by the wall (\(N\)) and speed of the block (\(v\))?
Choose graph (1), (2), (3), or (4).
✅ Answer & Explanation:
Correct Option: (1)
The normal reaction by the vertical wall provides the centripetal force: \(N = mv^2/R\).
Thus, \(N \propto v^2\). This is a parabolic relationship represented by graph (1).
The normal reaction by the vertical wall provides the centripetal force: \(N = mv^2/R\).
Thus, \(N \propto v^2\). This is a parabolic relationship represented by graph (1).
Q17. JEE Main 2022 (26 Jul Shift 1)
A monkey of mass 50 kg climbs on a rope which can withstand the tension (\(T\)) of 350 N. If monkey initially climbs down with an acceleration of \(4 \, \text{m/s}^2\) and then climbs up with an acceleration of \(5 \, \text{m/s}^2\). Choose the correct option : (\(g = 10 \, \text{m/s}^2\))
(1) \(T = 700 \, \text{N}\) while climbing upward (2) \(T = 350 \, \text{N}\) while going downward
(3) Rope will break while climbing upward (4) Rope will break while going downward
(1) \(T = 700 \, \text{N}\) while climbing upward (2) \(T = 350 \, \text{N}\) while going downward
(3) Rope will break while climbing upward (4) Rope will break while going downward
✅ Answer & Explanation:
Correct Option: (3)
- Climbing down: \(T = m(g - a) = 50(10 - 4) = 300 \, \text{N}\) (Safe).
- Climbing up: \(T = m(g + a) = 50(10 + 5) = 750 \, \text{N}\).
Since 750 N exceeds the limit of 350 N, the rope will break.
- Climbing down: \(T = m(g - a) = 50(10 - 4) = 300 \, \text{N}\) (Safe).
- Climbing up: \(T = m(g + a) = 50(10 + 5) = 750 \, \text{N}\).
Since 750 N exceeds the limit of 350 N, the rope will break.
Q18. JEE Main 2021 (20 Jul Shift 1)
A steel block of 10 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 mass of the iron cylinders are equal and of 20 kg each is (in N), [Take \(g = 10 \, \text{m/s}^2\) and \(\mu_s = 0.2\)]
(1) 716 (2) 686
(3) 714 (4) 684
(3) 714 (4) 684
✅ Answer & Explanation:
Correct Option: (2)
Total mass \(M = 10 + 3(20) = 70 \, \text{kg}\).
Equation of motion: \(Mg - R' = Ma\).
\(700 - R' = 70 \times 0.2 \implies R' = 700 - 14 = 686 \, \text{N}\).
Total mass \(M = 10 + 3(20) = 70 \, \text{kg}\).
Equation of motion: \(Mg - R' = Ma\).
\(700 - R' = 70 \times 0.2 \implies R' = 700 - 14 = 686 \, \text{N}\).
📘 Exam Preparation Tip for Laws of Motion
- Free Body Diagrams (FBD): Drawing accurate FBDs is the most important step. Identify all forces (gravity, normal, tension, friction) and their directions clearly.
- Newton's Second Law: Write equations \(F_{net} = ma\) separately for each block along different axes. For inclined planes, resolve forces into components parallel and perpendicular to the plane.
- Friction: Understand the difference between static and kinetic friction. Static friction adjusts to match the applied force up to a maximum value. Kinetic friction is constant and acts opposite to motion.
- Circular Motion: In uniform circular motion, remember that the net force towards the center provides the centripetal force: \(F_c = mv^2/r = m\omega^2 r\). This is not a separate force but the resultant of real forces.
- Pulley Problems: Assume a direction for acceleration and write equations for each block. Remember that tension is the same in a string passing over a massless, frictionless pulley, and acceleration is the same if the string is inextensible.
- Spring Force: Use Hooke's Law \(F = -kx\). Remember that the force is restoring and proportional to the displacement from the natural length.