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    the coefficient of static friction between block a of mass 2 kg

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    The coefficient of static friction, μs between block A of mass 2 kg

    The coefficient of static friction μs between block A of mass 2 kg and the table as shown in the figure is 0.2. What ... 0 kg (c) 0.2 kg (d) 0.4 kg

    The coefficient of static friction, μs between block A of mass 2 kg

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    स्रोत : www.sarthaks.com

    The coefficient of static friction, μ s, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. g =10 m / s 2A. 0.2 kgB. 0.4 kgC. 2.0 kgD. 4.0 kg

    The coefficient of static friction, μ s, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. g =10 m / s 2A. 0.2 kgB. 0.4 kgC. 2.0 kgD. 4.0 kg

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    The coefficient of static friction, μ s, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. g =10 m / s 2A. 0.2 kgB. 0.4 kgC. 2.0 kgD. 4.0 kg

    Question

    The coefficient of static friction,

    μ s , between block A

    of mass 2 kg and the table as shown in the figure is

    0.2.

    What would be the maximum mass value of block

    B

    so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless.

    ( g = 10 m/s 2 )

    A 2.0 kg B 4.0 kg C 0.2 kg D 0.4 kg Open in App Solution

    The correct option is B 0.4 kg

    Free body diagrams of two masses are

    We get equations T − f = m A a a n d f = μ N = μ m A g ⇒ T = m A a + μ m A g a n d T = m B g ( f o r a = 0 ) ∴ μ N = m B g ⇒ m B = μ m A = 0.2 × 2 = 0.4 kg .

    Hence, the correct answer is option (d).

    Suggest Corrections 3

    SIMILAR QUESTIONS

    Q. The coefficient of static friction is

    0.2 between block A of mass 2 kg

    and the table as shown in the figure. What would be the maximum mass value of block

    B

    so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless

    ( g = 10 m/s 2 )

    Q. The coefficient of static friction,

    μ s , between block A

    of mass 2 kg and the table as shown in the figure is

    0.2.

    What would be the maximum mass value of block

    B

    so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless.

    ( g = 10 m/s 2 )

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    स्रोत : byjus.com

    The coefficient of static friction, mus , between block A of mass 2 kg and the table as shown in the figure is 0.2 . What would be the maximum mass value of block B so that the two blocks do not move ? The string and the pulley are assumed to be smooth and massless. ( g = 10 m/s^2 )

    Click here👆to get an answer to your question ✍️ The coefficient of static friction, mus , between block A of mass 2 kg and the table as shown in the figure is 0.2 . What would be the maximum mass value of block B so that the two blocks do not move ? The string and the pulley are assumed to be smooth and massless. ( g = 10 m/s^2 )

    The coefficient of static friction, μ

    Question s ​

    , between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move ? The string and the pulley are assumed to be smooth and massless.

    (g=10m/s 2 )

    A

    4.0 kg

    B

    0.2 kg

    C

    0.4 kg

    D

    2.0 kg

    Hard Open in App Solution Verified by Toppr

    Correct option is C)

    The mass of the block A is 2kg, the coefficient of friction is 0.2.

    As the blocks do not move therefore, the tension in the string must be equal to the force of friction on block A.

    Applying Newton's second law for block A,

    T=μ s ​ m A ​ g =0.2×2×10 =4N

    Applying Newton's second law for block B,

    T=m B ​ g 4=m B ​ ×10 m B ​ =0.4kg

    Thus, the maximum mass of the block B is 0.4kg.

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    Mohammed 1 month ago
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