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Laws of Motion question

2023 · 31 Jan · Shift 2 · Q54
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  5. /2023 · 31 Jan · Shift 2 · Q54

Laws of Motion question

2023 · 31 Jan · Shift 2 · Q54

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A body of mass 10 kg10 \mathrm{~kg}10 kg is moving with an initial speed of 20 m/s20 \mathrm{~m} / \mathrm{s}20 m/s. The body stops after 5 s5 \mathrm{~s}5 s due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity g=10 ms−2g=10 \mathrm{~ms}^{-2}g=10 ms−2 )
  1. A
    0.3
  2. B
    0.2
  3. C
    0.5
  4. D
    0.4
View written solutionFree

Correct answer: D

  1. Given data

    • Mass of body: m=10 kgm = 10\,\text{kg}m=10kg
    • Initial speed: u=20 m/su = 20\,\text{m/s}u=20m/s
    • Final speed: v=0v = 0v=0 (body stops)
    • Time taken: t=5 st = 5\,\text{s}t=5s
    • Acceleration due to gravity: g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2

  2. Find the retardation

    Using the first equation of motion: v=u+atv = u + atv=u+at

    Substituting values: 0=20+a(5)0 = 20 + a(5)0=20+a(5) 5a=−205a = -205a=−20 a=−4 m/s2a = -4\,\text{m/s}^2a=−4m/s2

    So, the magnitude of retardation is: ∣a∣=4 m/s2|a| = 4\,\text{m/s}^2∣a∣=4m/s2

  3. Relate retardation to friction

    The only horizontal force stopping the body is friction.

    Friction force: f=μN=μmgf = \mu N = \mu mgf=μN=μmg

    By Newton's second law: f=maf = maf=ma

    Taking magnitudes: μmg=m∣a∣\mu mg = m|a|μmg=m∣a∣

    Cancel mmm: μg=∣a∣\mu g = |a|μg=∣a∣ μ=∣a∣g=410=0.4\mu = \frac{|a|}{g} = \frac{4}{10} = 0.4μ=g∣a∣​=104​=0.4

  4. Match with options

    μ=0.4\mu = 0.4μ=0.4

    Therefore, the correct option is D.

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