{"id":4588,"date":"2026-04-11T10:29:59","date_gmt":"2026-04-11T10:29:59","guid":{"rendered":"https:\/\/ksquareinstitute.in\/blog\/?p=4588"},"modified":"2026-04-11T10:34:32","modified_gmt":"2026-04-11T10:34:32","slug":"physics-laws-of-motion-quick-notes","status":"publish","type":"post","link":"https:\/\/ksquareinstitute.in\/blog\/physics-laws-of-motion-quick-notes\/","title":{"rendered":"Top PYQs from Laws of Motion with Concepts & Quick Notes for NEET"},"content":{"rendered":"\n

If you want one chapter in mechanics that can quickly improve your score, it is this one. Physics Laws of Motion Quick Notes<\/strong> is not just about Newton\u2019s laws. In NEET, this chapter blends force analysis, friction, tension, pseudo force ideas, circular motion basics, and connected-body reasoning into direct MCQs and numericals. The official NEET syllabus for Laws of Motion includes force and inertia, Newton\u2019s laws, momentum, impulse, friction, equilibrium of concurrent forces, and dynamics of uniform circular motion.<\/p>\n\n\n\n

This article is built as a dense revision resource using high-yield PYQ patterns adapted from NEET chapter-wise previous-year question banks and common exam trends<\/strong> for Laws of Motion.<\/p>\n\n\n\n

\"Physics<\/figure>\n\n\n\n

Physics Laws of Motion Quick Notes<\/h2>\n\n\n\n

Before jumping into the questions, revise these core points from Physics Laws of Motion Quick Notes<\/strong>:<\/p>\n\n\n\n

1. Newton\u2019s First Law<\/strong>
A body remains in rest or uniform motion unless acted upon by an external unbalanced force.<\/p>\n\n\n\n

2. Newton\u2019s Second Law<\/strong>F<\/mi>=<\/mo>m<\/mi>a<\/mi><\/mrow>F = ma<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

For variable momentum form:F<\/mi>=<\/mo>d<\/mi>p<\/mi><\/mrow>d<\/mi>t<\/mi><\/mrow><\/mfrac><\/mrow>F = \\frac{dp}{dt}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

3. Newton\u2019s Third Law<\/strong>
Every action has an equal and opposite reaction. These two forces act on different bodies.<\/p>\n\n\n\n

4. Linear Momentum<\/strong>p<\/mi>=<\/mo>m<\/mi>v<\/mi><\/mrow>p = mv<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

5. Impulse<\/strong>J<\/mi>=<\/mo>F<\/mi>\u0394<\/mi>t<\/mi>=<\/mo>\u0394<\/mi>p<\/mi><\/mrow>J = F\\Delta t = \\Delta p<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

6. Friction<\/strong><\/p>\n\n\n\n

    \n
  • Static friction: adjusts up to limiting value<\/li>\n<\/ul>\n\n\n\n

    f<\/mi>s<\/mi><\/msub>\u2264<\/mo>\u03bc<\/mi>s<\/mi><\/msub>N<\/mi><\/mrow>f_s \\leq \\mu_s N<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
      \n
    • Kinetic friction:<\/li>\n<\/ul>\n\n\n\n

      f<\/mi>k<\/mi><\/msub>=<\/mo>\u03bc<\/mi>k<\/mi><\/msub>N<\/mi><\/mrow>f_k = \\mu_k N<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      7. On an Inclined Plane<\/strong><\/p>\n\n\n\n

        \n
      • Along plane:<\/li>\n<\/ul>\n\n\n\n

        m<\/mi>g<\/mi>sin<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>mg\\sin\\theta<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
          \n
        • Normal reaction:<\/li>\n<\/ul>\n\n\n\n

          N<\/mi>=<\/mo>m<\/mi>g<\/mi>cos<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>N = mg\\cos\\theta<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

          8. Circular Motion<\/strong>F<\/mi>c<\/mi><\/msub>=<\/mo>m<\/mi>v<\/mi>2<\/mn><\/msup><\/mrow>r<\/mi><\/mfrac><\/mrow>F_c = \\frac{mv^2}{r}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

          9. Connected Bodies<\/strong>
          Treat the system together when possible. Internal tensions cancel in system equations.<\/p>\n\n\n\n

          10. Elevator Concept<\/strong><\/p>\n\n\n\n

            \n
          • Moving upward with acceleration a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math>a:<\/li>\n<\/ul>\n\n\n\n

            N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>+<\/mo>a<\/mi>)<\/mo><\/mrow>N = m(g+a)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n
            • Moving downward with acceleration a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math>a:<\/li>\n<\/ul>\n\n\n\n

              N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>\u2212<\/mo>a<\/mi>)<\/mo><\/mrow>N = m(g-a)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Now let us solve the 20 best questions<\/strong> in a dense, exam-useful format.<\/p>\n\n\n\n

              \n\n\n\n

              1) Force needed to produce acceleration<\/h2>\n\n\n\n

              A body of mass 4 kg is moving with acceleration 3 m\/s\u00b2. Find the net force acting on it.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Using Newton\u2019s second law:F<\/mi>=<\/mo>m<\/mi>a<\/mi>=<\/mo>4<\/mn>\u00d7<\/mo>3<\/mn>=<\/mo>12<\/mn> N<\/mtext><\/mrow>F = ma = 4 \\times 3 = 12\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 12 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> This is the most basic application of Physics Laws of Motion Quick Notes<\/strong>. Always check whether the question asks for net force or applied force.<\/p>\n\n\n\n

              \n\n\n\n

              2) Momentum change by constant force<\/h2>\n\n\n\n

              A force of 10 N acts on a body for 5 s. Find the change in momentum.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Impulse equals change in momentum:\u0394<\/mi>p<\/mi>=<\/mo>F<\/mi>\u0394<\/mi>t<\/mi>=<\/mo>10<\/mn>\u00d7<\/mo>5<\/mn>=<\/mo>50<\/mn> kg m\/s<\/mtext><\/mrow>\\Delta p = F\\Delta t = 10 \\times 5 = 50\\text{ kg m\/s}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 50 kg m\/s<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Impulse questions are direct favorites in NEET because they test the momentum form quickly.<\/p>\n\n\n\n

              \n\n\n\n

              3) Inertia-based concept<\/h2>\n\n\n\n

              Why does a passenger fall backward when a bus starts suddenly?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              When the bus starts suddenly, the lower part of the passenger\u2019s body moves forward with the bus due to contact force. The upper part tends to remain at rest because of inertia of rest.<\/p>\n\n\n\n

              Answer:<\/strong> Due to inertia of rest.<\/p>\n\n\n\n

              Concept:<\/strong> A very standard theory question from Physics Laws of Motion Quick Notes<\/strong>. Similar logic applies for falling forward when the bus stops suddenly.<\/p>\n\n\n\n

              \n\n\n\n

              4) Action-reaction pair<\/h2>\n\n\n\n

              A book rests on a table. Which of the following are an action-reaction pair?
              (a) Weight of book and normal on book
              (b) Normal on book and force by book on table
              (c) Weight of book and force by book on table<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Action-reaction forces act on different bodies.
              Normal on book is exerted by table on book. Equal and opposite force is exerted by book on table.<\/p>\n\n\n\n

              Answer: (b)<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Students often confuse balanced forces with action-reaction pairs. Balanced forces can act on the same body. Action-reaction never do.<\/p>\n\n\n\n

              \n\n\n\n

              5) Acceleration of two-block system<\/h2>\n\n\n\n

              Two blocks of masses 2 kg and 3 kg are connected by a light string on a frictionless horizontal surface. A horizontal force of 10 N is applied on the 2 kg block. Find acceleration of the system.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Treat both blocks as one system.<\/p>\n\n\n\n

              Total mass:M<\/mi>=<\/mo>2<\/mn>+<\/mo>3<\/mn>=<\/mo>5<\/mn> kg<\/mtext><\/mrow>M = 2+3=5\\text{ kg}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Acceleration:a<\/mi>=<\/mo>F<\/mi>M<\/mi><\/mfrac>=<\/mo>10<\/mn>5<\/mn><\/mfrac>=<\/mo>2<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>a = \\frac{F}{M}=\\frac{10}{5}=2\\text{ m\/s}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 m\/s\u00b2<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> In Physics Laws of Motion Quick Notes<\/strong>, always prefer the system approach first in connected-body problems.<\/p>\n\n\n\n

              \n\n\n\n

              6) Tension in the string<\/h2>\n\n\n\n

              Using the same system as above, find the tension in the string.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Take the 3 kg block. It is pulled only by tension.T<\/mi>=<\/mo>m<\/mi>a<\/mi>=<\/mo>3<\/mn>\u00d7<\/mo>2<\/mn>=<\/mo>6<\/mn> N<\/mtext><\/mrow>T = ma = 3 \\times 2 = 6\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 6 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> After finding acceleration of the system, take one block individually.<\/p>\n\n\n\n

              \n\n\n\n

              7) Block on rough horizontal surface<\/h2>\n\n\n\n

              A 5 kg block is pushed with a horizontal force of 30 N on a surface with coefficient of friction \u03bc<\/mi>=<\/mo>0.4<\/mn><\/mrow>\\mu = 0.4<\/annotation><\/semantics><\/math>\u03bc=0.4. Find acceleration. Take g<\/mi>=<\/mo>10<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>g=10\\text{ m\/s}^2<\/annotation><\/semantics><\/math>g=10 m\/s2.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Normal reaction:N<\/mi>=<\/mo>m<\/mi>g<\/mi>=<\/mo>5<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>50<\/mn> N<\/mtext><\/mrow>N = mg = 5\\times10=50\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Friction:f<\/mi>=<\/mo>\u03bc<\/mi>N<\/mi>=<\/mo>0.4<\/mn>\u00d7<\/mo>50<\/mn>=<\/mo>20<\/mn> N<\/mtext><\/mrow>f = \\mu N = 0.4\\times50=20\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Net force:F<\/mi>n<\/mi>e<\/mi>t<\/mi><\/mrow><\/msub>=<\/mo>30<\/mn>\u2212<\/mo>20<\/mn>=<\/mo>10<\/mn> N<\/mtext><\/mrow>F_{net}=30-20=10\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Acceleration:a<\/mi>=<\/mo>10<\/mn>5<\/mn><\/mfrac>=<\/mo>2<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>a=\\frac{10}{5}=2\\text{ m\/s}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 m\/s\u00b2<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> This is one of the most repeated formats in Physics Laws of Motion Quick Notes<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              8) Limiting friction<\/h2>\n\n\n\n

              A body of mass 10 kg is on a rough horizontal surface with coefficient of static friction 0.3<\/mn><\/mrow>0.3<\/annotation><\/semantics><\/math>0.3. What is the maximum static friction? Take g<\/mi>=<\/mo>10<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>g=10\\text{ m\/s}^2<\/annotation><\/semantics><\/math>g=10 m\/s2.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              N<\/mi>=<\/mo>m<\/mi>g<\/mi>=<\/mo>10<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>100<\/mn> N<\/mtext><\/mrow>N = mg = 10\\times10=100\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Maximum static friction:f<\/mi>s<\/mi>,<\/mo>max<\/mi>\u2061<\/mo><\/mrow><\/msub>=<\/mo>\u03bc<\/mi>s<\/mi><\/msub>N<\/mi>=<\/mo>0.3<\/mn>\u00d7<\/mo>100<\/mn>=<\/mo>30<\/mn> N<\/mtext><\/mrow>f_{s,\\max}=\\mu_s N = 0.3 \\times 100 = 30\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 30 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Static friction is self-adjusting. It equals the applied force only up to 30 N here.<\/p>\n\n\n\n

              \n\n\n\n

              9) Block on incline without friction<\/h2>\n\n\n\n

              A block of mass 2 kg is placed on a smooth incline of angle 30<\/mn>\u2218<\/mo><\/msup><\/mrow>30^\\circ<\/annotation><\/semantics><\/math>30\u2218. Find acceleration down the plane.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              For smooth incline:a<\/mi>=<\/mo>g<\/mi>sin<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>a = g\\sin\\theta<\/annotation><\/semantics><\/math>a<\/mi>=<\/mo>10<\/mn>\u00d7<\/mo>sin<\/mi>\u2061<\/mo>30<\/mn>\u2218<\/mo><\/msup>=<\/mo>10<\/mn>\u00d7<\/mo>1<\/mn>2<\/mn><\/mfrac>=<\/mo>5<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>a = 10 \\times \\sin30^\\circ = 10 \\times \\frac{1}{2}=5\\text{ m\/s}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 5 m\/s\u00b2<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> This is a must-remember direct result in Physics Laws of Motion Quick Notes<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              10) Normal reaction on incline<\/h2>\n\n\n\n

              For the same block, find normal reaction.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              N<\/mi>=<\/mo>m<\/mi>g<\/mi>cos<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>N = mg\\cos\\theta<\/annotation><\/semantics><\/math>N<\/mi>=<\/mo>2<\/mn>\u00d7<\/mo>10<\/mn>\u00d7<\/mo>cos<\/mi>\u2061<\/mo>30<\/mn>\u2218<\/mo><\/msup>=<\/mo>20<\/mn>\u00d7<\/mo>3<\/mn><\/msqrt>2<\/mn><\/mfrac>=<\/mo>10<\/mn>3<\/mn><\/msqrt> N<\/mtext><\/mrow>N = 2\\times10\\times\\cos30^\\circ = 20\\times\\frac{\\sqrt3}{2}=10\\sqrt3\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 10310\\sqrt3103\u200b N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Along incline use m<\/mi>g<\/mi>sin<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>mg\\sin\\theta<\/annotation><\/semantics><\/math>mgsin\u03b8, perpendicular use m<\/mi>g<\/mi>cos<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>mg\\cos\\theta<\/annotation><\/semantics><\/math>mgcos\u03b8.<\/p>\n\n\n\n
              \n\n\n\n

              11) Lift moving upward<\/h2>\n\n\n\n

              A man of mass 60 kg stands on a weighing machine in a lift accelerating upward at 2 m\/s\u00b2. What will be the reading? Take g<\/mi>=<\/mo>10<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>g=10\\text{ m\/s}^2<\/annotation><\/semantics><\/math>g=10 m\/s2.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Apparent weight:N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>+<\/mo>a<\/mi>)<\/mo>=<\/mo>60<\/mn>(<\/mo>10<\/mn>+<\/mo>2<\/mn>)<\/mo>=<\/mo>60<\/mn>\u00d7<\/mo>12<\/mn>=<\/mo>720<\/mn> N<\/mtext><\/mrow>N = m(g+a)=60(10+2)=60\\times12=720\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 720 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Elevator questions are extremely scoring from Physics Laws of Motion Quick Notes<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              12) Lift moving downward<\/h2>\n\n\n\n

              If the same lift moves downward with acceleration 2 m\/s\u00b2, find the reading.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>\u2212<\/mo>a<\/mi>)<\/mo>=<\/mo>60<\/mn>(<\/mo>10<\/mn>\u2212<\/mo>2<\/mn>)<\/mo>=<\/mo>60<\/mn>\u00d7<\/mo>8<\/mn>=<\/mo>480<\/mn> N<\/mtext><\/mrow>N = m(g-a)=60(10-2)=60\\times8=480\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 480 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Weight becomes smaller in downward acceleration.<\/p>\n\n\n\n

              \n\n\n\n

              13) Weightlessness condition<\/h2>\n\n\n\n

              Under what condition will the weighing machine read zero in a lift?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              The reading becomes zero when:N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>\u2212<\/mo>a<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>N = m(g-a)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              So,a<\/mi>=<\/mo>g<\/mi><\/mrow>a=g<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer:<\/strong> When the lift accelerates downward with acceleration equal to g<\/mi><\/mrow>g<\/annotation><\/semantics><\/math>g.<\/p>\n\n\n\n

              Concept:<\/strong> This is the weightlessness condition.<\/p>\n\n\n\n

              \n\n\n\n

              14) Conservation of momentum<\/h2>\n\n\n\n

              A 2 kg body moving with 6 m\/s collides with a 4 kg body at rest and sticks to it. Find common velocity after collision.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Since bodies stick together, collision is perfectly inelastic.<\/p>\n\n\n\n

              Initial momentum:p<\/mi>i<\/mi><\/msub>=<\/mo>2<\/mn>\u00d7<\/mo>6<\/mn>+<\/mo>4<\/mn>\u00d7<\/mo>0<\/mn>=<\/mo>12<\/mn><\/mrow>p_i = 2\\times6 + 4\\times0 = 12<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Final momentum:(<\/mo>2<\/mn>+<\/mo>4<\/mn>)<\/mo>v<\/mi>=<\/mo>6<\/mn>v<\/mi><\/mrow>(2+4)v = 6v<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Using conservation of momentum:6<\/mn>v<\/mi>=<\/mo>12<\/mn>\u21d2<\/mo>v<\/mi>=<\/mo>2<\/mn> m\/s<\/mtext><\/mrow>6v=12 \\Rightarrow v=2\\text{ m\/s}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 m\/s<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Very common momentum question in Physics Laws of Motion Quick Notes<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              15) Force between two interacting bodies<\/h2>\n\n\n\n

              A 5 kg block pushes a 10 kg block with force 20 N. What is the force exerted by the 10 kg block on the 5 kg block?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              By Newton\u2019s third law, the force is equal in magnitude and opposite in direction.<\/p>\n\n\n\n

              Answer: 20 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Third-law forces are always equal, regardless of masses.<\/p>\n\n\n\n

              \n\n\n\n

              16) Circular motion on level road<\/h2>\n\n\n\n

              A car of mass 1000 kg moves in a circular path of radius 50 m with speed 10 m\/s. Find centripetal force.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              F<\/mi>c<\/mi><\/msub>=<\/mo>m<\/mi>v<\/mi>2<\/mn><\/msup><\/mrow>r<\/mi><\/mfrac><\/mrow>F_c = \\frac{mv^2}{r}<\/annotation><\/semantics><\/math> F<\/mi>c<\/mi><\/msub>=<\/mo>1000<\/mn>\u00d7<\/mo>10<\/mn>2<\/mn><\/msup><\/mrow>50<\/mn><\/mfrac>=<\/mo>1000<\/mn>\u00d7<\/mo>100<\/mn><\/mrow>50<\/mn><\/mfrac>=<\/mo>2000<\/mn> N<\/mtext><\/mrow>F_c = \\frac{1000\\times10^2}{50}=\\frac{1000\\times100}{50}=2000\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2000 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Circular motion is officially included under Laws of Motion in NEET.<\/p>\n\n\n\n

              \n\n\n\n

              17) Minimum force to move a block<\/h2>\n\n\n\n

              A 4 kg block lies on a rough horizontal surface with coefficient of static friction 0.5. Find the minimum horizontal force required to just move it. Take g<\/mi>=<\/mo>10<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>g=10\\text{ m\/s}^2<\/annotation><\/semantics><\/math>g=10 m\/s2.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Normal reaction:N<\/mi>=<\/mo>m<\/mi>g<\/mi>=<\/mo>4<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>40<\/mn> N<\/mtext><\/mrow>N=mg=4\\times10=40\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Maximum static friction:f<\/mi>s<\/mi>,<\/mo>max<\/mi>\u2061<\/mo><\/mrow><\/msub>=<\/mo>\u03bc<\/mi>s<\/mi><\/msub>N<\/mi>=<\/mo>0.5<\/mn>\u00d7<\/mo>40<\/mn>=<\/mo>20<\/mn> N<\/mtext><\/mrow>f_{s,\\max}=\\mu_s N=0.5\\times40=20\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              So the least force required to start motion is 20 N<\/strong>.<\/p>\n\n\n\n

              Answer: 20 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> \u201cJust move\u201d means limiting friction.<\/p>\n\n\n\n

              \n\n\n\n

              18) Retardation due to friction only<\/h2>\n\n\n\n

              A 2 kg block slides on a rough horizontal surface with coefficient of kinetic friction 0.2. Find retardation. Take g<\/mi>=<\/mo>10<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>g=10\\text{ m\/s}^2<\/annotation><\/semantics><\/math>g=10 m\/s2.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Friction:f<\/mi>k<\/mi><\/msub>=<\/mo>\u03bc<\/mi>k<\/mi><\/msub>N<\/mi>=<\/mo>\u03bc<\/mi>k<\/mi><\/msub>m<\/mi>g<\/mi>=<\/mo>0.2<\/mn>\u00d7<\/mo>2<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>4<\/mn> N<\/mtext><\/mrow>f_k = \\mu_k N = \\mu_k mg = 0.2\\times2\\times10=4\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Retardation:a<\/mi>=<\/mo>f<\/mi>m<\/mi><\/mfrac>=<\/mo>4<\/mn>2<\/mn><\/mfrac>=<\/mo>2<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>a = \\frac{f}{m} = \\frac{4}{2}=2\\text{ m\/s}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 m\/s\u00b2<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> On a horizontal rough surface, retardation due to friction is simply:a<\/mi>=<\/mo>\u03bc<\/mi>g<\/mi><\/mrow>a=\\mu g<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n\n\n\n

              19) Man pulling a rope<\/h2>\n\n\n\n

              A man pulls a rope tied to a wall with force 100 N. What force does the wall exert on the man through the rope?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              The rope transmits tension equal to applied pull if massless and in equilibrium.<\/p>\n\n\n\n

              So tension in rope = 100 N.
              The wall pulls back through the rope with the same tension.<\/p>\n\n\n\n

              Answer: 100 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Many students overcomplicate this. Tension is the key idea.<\/p>\n\n\n\n

              \n\n\n\n

              20) Net force from velocity change<\/h2>\n\n\n\n

              A body of mass 2 kg changes its velocity from 5 m\/s to 15 m\/s in 2 s. Find net force.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Acceleration:a<\/mi>=<\/mo>v<\/mi>\u2212<\/mo>u<\/mi><\/mrow>t<\/mi><\/mfrac>=<\/mo>15<\/mn>\u2212<\/mo>5<\/mn><\/mrow>2<\/mn><\/mfrac>=<\/mo>5<\/mn> m\/s<\/mtext>2<\/mn><\/msup><\/mrow>a=\\frac{v-u}{t}=\\frac{15-5}{2}=5\\text{ m\/s}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Force:F<\/mi>=<\/mo>m<\/mi>a<\/mi>=<\/mo>2<\/mn>\u00d7<\/mo>5<\/mn>=<\/mo>10<\/mn> N<\/mtext><\/mrow>F=ma=2\\times5=10\\text{ N}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 10 N<\/strong><\/p>\n\n\n\n

              Concept:<\/strong> Kinematics plus Newton\u2019s second law is a classic NEET combination.<\/p>\n\n\n\n

              \n\n\n\n

              Quick Revision Box: Physics Laws of Motion Quick Notes<\/h1>\n\n\n\n

              Use this as your final 1-minute recap:<\/p>\n\n\n\n

              \u2192 p<\/mi>=<\/mo>m<\/mi>v<\/mi><\/mrow>p=mv<\/annotation><\/semantics><\/math>\u2192 J<\/mi>=<\/mo>F<\/mi>\u0394<\/mi>t<\/mi>=<\/mo>\u0394<\/mi>p<\/mi><\/mrow>J=F\\Delta t=\\Delta p<\/annotation><\/semantics><\/math>\u2192 f<\/mi>s<\/mi><\/msub>\u2264<\/mo>\u03bc<\/mi>s<\/mi><\/msub>N<\/mi><\/mrow>f_s \\le \\mu_s N<\/annotation><\/semantics><\/math>\u2192 f<\/mi>k<\/mi><\/msub>=<\/mo>\u03bc<\/mi>k<\/mi><\/msub>N<\/mi><\/mrow>f_k = \\mu_k N<\/annotation><\/semantics><\/math>\u2192 N<\/mi>=<\/mo>m<\/mi>g<\/mi><\/mrow>N = mg<\/annotation><\/semantics><\/math>\u2192 N<\/mi>=<\/mo>m<\/mi>g<\/mi>cos<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mrow>N = mg\\cos\\theta<\/annotation><\/semantics><\/math>\u2192 a<\/mi>=<\/mo>g<\/mi>sin<\/mi>\u2061<\/mo>\u03b8<\/mi><\/mspace>(smooth incline)<\/mtext><\/mrow>a = g\\sin\\theta \\quad \\text{(smooth incline)}<\/annotation><\/semantics><\/math>\u2192 F<\/mi>c<\/mi><\/msub>=<\/mo>m<\/mi>v<\/mi>2<\/mn><\/msup><\/mrow>r<\/mi><\/mfrac><\/mrow>F_c=\\frac{mv^2}{r}<\/annotation><\/semantics><\/math>\u2192 N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>+<\/mo>a<\/mi>)<\/mo><\/mspace>(lift upward)<\/mtext><\/mrow>N = m(g+a) \\quad \\text{(lift upward)}<\/annotation><\/semantics><\/math>\u2192 N<\/mi>=<\/mo>m<\/mi>(<\/mo>g<\/mi>\u2212<\/mo>a<\/mi>)<\/mo><\/mspace>(lift downward)<\/mtext><\/mrow>N = m(g-a) \\quad \\text{(lift downward)}<\/annotation><\/semantics><\/math>\u2192 <\/p>\n\n\n\n

              These formulas form the heart of Physics Laws of Motion Quick Notes<\/strong> and cover a major chunk of NEET-level questions from this chapter.<\/p>\n\n\n\n

              Most Important Exam Patterns from This Chapter<\/h1>\n\n\n\n

              In NEET, Laws of Motion questions are commonly asked from:<\/p>\n\n\n\n

                \n
              • Newton\u2019s laws and inertia-based concepts<\/li>\n\n\n\n
              • Tension in connected systems<\/li>\n\n\n\n
              • Static and kinetic friction<\/li>\n\n\n\n
              • Inclined plane force analysis<\/li>\n\n\n\n
              • Lift problems and apparent weight<\/li>\n\n\n\n
              • Momentum conservation<\/li>\n\n\n\n
              • Impulse<\/li>\n\n\n\n
              • Centripetal force basics<\/li>\n<\/ul>\n\n\n\n

                The topic spread matches the current NEET Laws of Motion syllabus and chapter-wise PYQ collections.<\/p>\n\n\n\n

                Final Takeaway<\/h1>\n\n\n\n

                If you revise Physics Laws of Motion Quick Notes<\/strong> properly, this chapter becomes one of the easiest scoring areas in mechanics. The trick is not memorizing random answers. The trick is recognizing which model the question belongs to: force balance, friction, incline, momentum, tension, or circular motion. Once that is clear, the solution becomes fast.<\/p>\n\n\n\n

                For NEET revision, do these 20 questions twice. First for understanding, second for speed. That alone can make this chapter feel much lighter.<\/p>\n\n\n\n

                If you want, next I can turn this into a full SEO blog format with meta title, meta description, FAQ, and dense intro\/outro<\/strong> matching your usual article style.<\/p>\n","protected":false},"excerpt":{"rendered":"

                If you want one chapter in mechanics that can quickly improve your score, it is this one. Physics Laws of Motion Quick Notes is not just about Newton\u2019s laws. In NEET, this chapter blends force analysis, friction, tension, pseudo force ideas, circular motion basics, and connected-body reasoning into direct MCQs and numericals. The official NEET […]<\/p>\n","protected":false},"author":1,"featured_media":4589,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70,2],"tags":[1069,1074,1065,1068,1067,1063,1064,1070,1072,855,1073,858,1062,1071,1066],"class_list":["post-4588","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-physics","category-neet","tag-circular-motion-basics","tag-force-and-motion-questions","tag-friction-questions-neet","tag-impulse-and-momentum-neet","tag-inclined-plane-questions","tag-laws-of-motion-neet-pyqs","tag-laws-of-motion-numericals","tag-laws-of-motion-revision-notes","tag-mechanics-chapter-neet","tag-neet-physics-mechanics","tag-neet-physics-questions-with-solutions","tag-newton-laws-questions","tag-physics-laws-of-motion-quick-notes","tag-physics-pyq-practice","tag-tension-problems-physics"],"blocksy_meta":{"page_structure_type":"type-1","styles_descriptor":{"styles":{"desktop":"","tablet":"","mobile":""},"google_fonts":[],"version":6}},"_links":{"self":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4588","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/comments?post=4588"}],"version-history":[{"count":4,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4588\/revisions"}],"predecessor-version":[{"id":4595,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4588\/revisions\/4595"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media\/4589"}],"wp:attachment":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media?parent=4588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/categories?post=4588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/tags?post=4588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}