Laws of Motion

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  1. A block B is pushed momentarily along a horizontal surface with an initial velocity V. If is the coefficient of sliding friction between B and the surface, block B will come to rest after a time








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    Friction is the retarding force for the block
    F = ma = µR = µmg
    Therefore, from the first equation of motion
    v = u – at
    0 = V – µg × t ⇒ V/μg = t

    Correct Option: D

    Friction is the retarding force for the block
    F = ma = µR = µmg
    Therefore, from the first equation of motion
    v = u – at
    0 = V – µg × t ⇒ V/μg = t

  1. The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by








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    For upper half of inclined plane
    v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ
    For lower half of inclined plane
    0 = u² + 2 g (sin θ – µ cos θ) S/2
    ⇒ – gS sin θ = gS ( sinθ – µ cos θ)
    ⇒ 2 sin θ = µ cos θ
    ⇒ µ= 2sinθ/cosθ = 2 tan θ

    Correct Option: B


    For upper half of inclined plane
    v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ
    For lower half of inclined plane
    0 = u² + 2 g (sin θ – µ cos θ) S/2
    ⇒ – gS sin θ = gS ( sinθ – µ cos θ)
    ⇒ 2 sin θ = µ cos θ
    ⇒ µ= 2sinθ/cosθ = 2 tan θ

  1. A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is µ = 0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 ms–2, is








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    Frictional force on the box f = µmg
    ∴ Acceleration in the box
    a = µg = 5 ms–2
    v² = u² + 2as
    ⇒ 0 = 2² + 2 × (5) s
    ⇒ s = – 2/5 w.r.t. belt
    ⇒ distance = 0.4 m

    Correct Option: D

    Frictional force on the box f = µmg
    ∴ Acceleration in the box
    a = µg = 5 ms–2
    v² = u² + 2as
    ⇒ 0 = 2² + 2 × (5) s
    ⇒ s = – 2/5 w.r.t. belt
    ⇒ distance = 0.4 m

  1. A block of mass m is in contact with the cart C as shown in the Figure.

    The coefficient of static friction between the block and the cart is . The acceleration of the cart that will prevent the block from falling satisfies:








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    Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force mα due to acceleration

    ∴ N = mα
    For the block not to fall, frictional force, Ff ≥ mg
    ⇒ μN ≥ mg
    ⇒ μmα ≥ mg
    ⇒ α ≥ g/μ

    Correct Option: C

    Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force mα due to acceleration

    ∴ N = mα
    For the block not to fall, frictional force, Ff ≥ mg
    ⇒ μN ≥ mg
    ⇒ μmα ≥ mg
    ⇒ α ≥ g/μ

  1. A block of mass 1 kg is placed on a truck which accelerates with acceleration 5m/s². The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is








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    Maximum friction force = µmg
    = .6 × 1 × 9.8 = 5.88 N
    But here required friction force
    = ma = 1 × 5 = 5 N

    Correct Option: A

    Maximum friction force = µmg
    = .6 × 1 × 9.8 = 5.88 N
    But here required friction force
    = ma = 1 × 5 = 5 N