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A 60kg man is running on horizontal road...

A 60kg man is running on horizontal road with acceleration `2ms^(-1)` friction coefficient between foot of the man and the ground is `mu=0.4` then

A

Frictional force acting on the man is 120N

B

Frictional force acting on the man is 240N

C

Work done by friction on the man during his 2m displacement will be 240J

D

Direction of the frictional force acting on the man is opposite to his direction of motion.

Text Solution

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f=ma
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Knowledge Check

  • A vehicle of mass M is moving on a rough horizontal road with a momentum P. If the coefficient of friction between the tyres and the road is mu , then the stopping distance is

    A
    `(P)/(2muMg)`
    B
    `(P^(2))/(2muMg)`
    C
    `(P^(2))/(2muM^(2)g)`
    D
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    A
    `(P)/(2muMg)`
    B
    `(P^(2))/(2muMg)`
    C
    `(P^(2))/(2muM^(2)g)`
    D
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  • A man is running on the ground .It is known that the coefficient of friction between the man and the ground is mu . Then which of the following statements is correct

    A
    Normal reaction between man and ground is equal to weight of man
    B
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    C
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    D
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