Laws of Motion

Laws of Motion

Easy Questions
Moderate Questions
Difficult Questions

Mechanics

Physical World, Units and Measurements
Motion in a Straight Line
Motion in a Plane
Laws of Motion
Work, Energy and Power
Gravitation
Mechanical Properties of Solids
Mechanical Properties of Fluids
System of Particles and Rotational Motion
  1. A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is µk. When the block A is sliding on the table, the tension in the string is








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: D

    NA

  1. Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g = 10 m/s²)








  1. View Hint View Answer Discuss in Forum

    Here u = 72 km/h = 20 m/s; v = 0;
    a = – µg = – 0.5 × 10 = – 5 m/s²
    As v² = u² + 2as,

    ∴ s =
    (v² - u²)
    		 =
    (0 - (20)²)
    		 = 40 m
    2a2 × (-5)

    Correct Option: B

    Here u = 72 km/h = 20 m/s; v = 0;
    a = – µg = – 0.5 × 10 = – 5 m/s²
    As v² = u² + 2as,

    ∴ s =
    (v² - u²)
    		 =
    (0 - (20)²)
    		 = 40 m
    2a2 × (-5)

  1. A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is








  1. View Hint View Answer Discuss in Forum

    The force of friction on the chain lying on the table should be equal to the weight of the hanging chain. Let
    ρ = mass per unit length of the chain
    µ = coefficient of friction
    l = length of the total chain
    x = length of hanging chain
    Now, µ(l – x) ρg = xρg or µ(l – x) = x
    or µl = (µ + 1)x or x = µl/(µ + 1)

    ∴ x =
    0.25l
    		 =
    0.25l
    		 = 0.2l
    (0.25 + 1)1.25

    x
    		 = 0.2 = 20%
    l

    Correct Option: A

    The force of friction on the chain lying on the table should be equal to the weight of the hanging chain. Let
    ρ = mass per unit length of the chain
    µ = coefficient of friction
    l = length of the total chain
    x = length of hanging chain
    Now, µ(l – x) ρg = xρg or µ(l – x) = x
    or µl = (µ + 1)x or x = µl/(µ + 1)

    ∴ x =
    0.25l
    		 =
    0.25l
    		 = 0.2l
    (0.25 + 1)1.25

    x
    		 = 0.2 = 20%
    l

  1. Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is








  1. View Hint View Answer Discuss in Forum

    In presence of friction a = (g sinθ – µg cos θ)
    ∴ Time taken to slide down the plane

    t1 = √
    2s
    		 = √
    2s
    		 = 48
    ag(sinθ - μ cosθ)

    In absence of friction t2 = √
    2s
    g sinθ

    t1 2t2
    &there; t1² 4t2²
    or =
    2s
    		 =
    2s × 4
    g(sinθ - μcosθ)gsinθ

    sin θ = 4 sinθ – 4µ cos θ
    μ =
    3
    		 tanθ =
    3
    		 = 0.75
    44


    Correct Option: B

    In presence of friction a = (g sinθ – µg cos θ)
    ∴ Time taken to slide down the plane

    t1 = √
    2s
    		 = √
    2s
    		 = 48
    ag(sinθ - μ cosθ)

    In absence of friction t2 = √
    2s
    g sinθ

    t1 2t2
    &there; t1² 4t2²
    or =
    2s
    		 =
    2s × 4
    g(sinθ - μcosθ)gsinθ

    sin θ = 4 sinθ – 4µ cos θ
    μ =
    3
    		 tanθ =
    3
    		 = 0.75
    44


  1. A car is negotiating a curved road of radius R. The road is banked at an angle θ. the coefficient of friction between the tyres of the car and the road is µs. The maximum safe velocity on this road is :








  1. View Hint View Answer Discuss in Forum

    On a banked road,

    Vmax²
    		 =
    μs + tanθ
    Rg1 - μs tanθ

    Maximum safe velocity of a car on the banked road
    Vmax = √ Rg
    μs + tanθ
    1 - μs + tanθ

    Correct Option: B

    On a banked road,

    Vmax²
    		 =
    μs + tanθ
    Rg1 - μs tanθ

    Maximum safe velocity of a car on the banked road
    Vmax = √ Rg
    μs + tanθ
    1 - μs + tanθ