two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg

Question

two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg

and the incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the

incline is µ2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10m/s^2).​

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Thu Nguyệt 5 months 2021-08-18T15:14:38+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-18T15:15:39+00:00

    Explanation:

    2.7m/s2

    I hope its helpful

    0
    2021-08-18T15:16:24+00:00

    Answer:

    The acceleration of 2.0 kg block is 2.7 m/s²

    Explanation:

    Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

    Force down the plane on the system

    = (4 + 2) g sin30°

    = (6)(10)(½)

    = 30N

    Force up the plane on the system

    = µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°

    = (2µ₁ + 4µ₂) g cos30°

    = (2 × 0.2 + 4 × 0.3)(10)(√3/2)

    ≈ 13.76 N

    ∴ Net force down the plane is F

    F = 30 – 13.76

    F = 16.24 N

    ∴Acceleration of both the blocks down the

    plane will b a

    a = F ÷ (4 + 2)

    a = 16.24 ÷ 6

    a = 2.7 m/s²

    Thus, The acceleration of 2.0 kg block is

    2.7 m/s²

    -TheUnknownScientist

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