Pavati Grover
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A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

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NCERT Solutions for Class 9 Science Chapter 9
Force and Laws of Motion
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Exercises Questions
Page No-128
Questions No-6

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1 Answer

  1. Initial velocity of the stone, u = 20 m/s
    Final velocity of the stone, v = 0 m/s
    Distance covered by the stone, s = 50 m
    According to the third equation of motion:
    v² = u² + 2as
    Where,
    Acceleration, a
    (0)2 = (20)² + 2 × a × 50
    a = – 4 m/s²
    The negative sign indicates that acceleration is acting against the motion of the
    stone.
    Mass of the stone, m = 1 kg
    From Newton’s second law of motion:
    Force,
    F = Mass x Acceleration
    F= ma
    F= 1 × (– 4) = – 4 N
    Hence, the force of friction between the stone and the ice is – 4 N.

    For more answers visit to website:
    https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/

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