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    a book of weight 20n is pressed between two hands and each hand exerts a force of 40n. if the book just starts to slide down. coefficient of friction is

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    A book of weight 20 N is pressed between two hands and each hand exerts a force of 40 N . If the book just starts to slide down, then the coefficient of friction is

    Click here👆to get an answer to your question ✍️ A book of weight 20 N is pressed between two hands and each hand exerts a force of 40 N . If the book just starts to slide down, then the coefficient of friction is

    Question

    A book of weight 20 N is pressed between two hands and each hand exerts a force of 40 N. If the book just starts to slide down, then the coefficient of friction is

    A

    0.25

    B

    0.2

    C

    0.5

    D

    0.1

    Medium Open in App Solution Verified by Toppr

    Correct option is A)

    We know:

    frictional force =μ x normal force

    Now,

    The book is pressed by two 40 N force acting normal, and the book tends to slide down,

    Thus, frictional force =2×μN

    ∴2×40μ=20⇒μ=0.25

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    A book of weight 20N is pressed between two hands and class 11 physics CBSE

    A book of weight 20N is pressed between two hands and each hand exerts a force of 40 N If the book just starts to slide down then the coefficient of friction is A 025 B 02 C 05 D 01

    A book of weight 20N is pressed between two hands and each hand exerts a force of 40 N. If the book just starts to slide down, then the coefficient of friction is:

    A) 0.25 B) 0.2 C) 0.5. D) 0.1. Answer Verified 212.1k+ views

    Hint:Friction is defined as the resistance to the motion of the body the coefficient of friction which works on the body is kinematic coefficient of friction if the body is in motion and the friction is static coefficient of friction if the body is at rest.Formula used:The formula of the friction force is given by,

    f s =μ⋅N fs=μ⋅N Where μ μ

    is the coefficient of friction and

    N N

    is the normal reaction.

    The formula of weight is given by,

    w=mg w=mg

    Where m is the mass of the body and g is the acceleration due to gravity.

    Complete step-by-step answer:

    It is given in the problem that a book is held by two hands which weighs 20N and the applied force by each hand is 40N and we need to find the value of force for the condition when the book starts to slip down to the ground.

    Here the friction force will be equal to the weight of the book,

    ⇒ f s =w ⇒fs=w

    As the formula of the friction is given by,

    f s =μ⋅N fs=μ⋅N Where μ μ

    is the coefficient of friction and

    N N

    is the normal reaction.

    Also the formula of weight is given by,

    w=mg w=mg

    Where m is the mass of the body and g is the acceleration due to gravity.

    ⇒μ⋅N=mg ⇒μ⋅N=mg

    As the force applied by each hand is equal and the weight of the body is 20N.

    ⇒μ⋅(2×40)=20 ⇒μ⋅(2×40)=20 ⇒μ= 20 (2×40) ⇒μ=20(2×40) ⇒μ= 1 4 ⇒μ=14 ⇒μ=0⋅25 ⇒μ=0⋅25

    The coefficient of friction is equal to

    μ=0⋅25 μ=0⋅25 .

    The correct answer for this problem is option A.Note:The formula of the friction is dependent on the coefficient of friction and normal reaction and the value of normal reaction is not always equal to the weight of the body; sometimes the applied force can also become the normal force as in this problem.

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    A book of weight 20N is pressed between two hands and each hand exerts a force of 40N. If the block just starts to slide down Coefficient of friction is .

    f(s) = mu(s) N ,mu = (f(s))/(N) = (mg)/(2F)A book of weight 20N is pressed between two hands and each hand exerts a force of 40N. If the block just starts to slide down Coefficient of friction is .

    Home > English > Class 11 > Physics > Chapter >

    Newtons Laws Of Motion

    >

    A book of weight 20N is presse...

    A book of weight 20N 20N

    is pressed between two hands and each hand exerts a force of

    40N 40N

    . If the block just starts to slide down Coefficient of friction is .

    Updated On: 27-06-2022

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    Text Solution Open Answer in App A 0.25 0.25 B 0.2 0.2 C 0.5 0.5 D 0.1 0.1 Answer

    The correct Answer is A

    Solution f s = μ s N,μ= f s N = mg 2F fs=μsN,μ=fsN=mg2F Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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    Mohammed 13 day ago
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