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A block of mass M is kept on a frictionl...

A block of mass M is kept on a frictionless horizontal table. A body of mass m is kept on the block. If a force F, Parallel to the table top is applied on the block, the body on the block tends to slide backwards. What is the coefficient of static friction between the block and the body?

Text Solution

Verified by Experts

The correct Answer is:
`(F)/((M+m)g)`
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Knowledge Check

  • A block of mass m_(2) is placed on a horizontal table and another block of mass m_(1) is placed on top of it. An increasing horizontal force F = at is exerted on the upper block but the lower block never moves as a result. If the co-efficient of friction between the blocks is mu_(1) and that between the lower block and the table is mu_(2) , then what is the maximum possible value of mu_(1)//mu_(2) ?

    A
    `m_(2)/(m_(1))`
    B
    `1+(m_(2))/(m_(1))`
    C
    `(m_(1))/(m_(2))`
    D
    `1+(m_(1))/(m_(2))`

  • A block A of mass m_(1) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m_(2) is suspended. The coefficient of kinetic friction between the block and the table is mu_(k) . When the block A is sliding on the table the tension in the string is

    A
    `((m_(2)+mu_(k)m_(1))g)/((m_(1)+m_(2)))`
    B
    `((m_(2)-mu_(k)m_(1))g)/((m_(1)+m_(2)))`
    C
    `(m_(1)m_(2)(1+mu_(k))g)/((m_(1)+m_(2)))`
    D
    `(m_(1)m_(2)(1-mu_(k))g)/((m_(1)+m_(2)))`

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