A body of mass $20 k g$ is kept initially at rest. A force of $80 N$ is applied on the body then the acceleration produced in the body is $3 m / s^{2}$ , force of friction acting on the body is

80 N

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12 N

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20 N

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Zero

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Solution

The correct option is C $20 N$
Mass of the body, $m = 20 k g$
Applied force, $F = 80 N$
Acceleration produced, $a = 3 m / s^{2}$
If force $F^{'}$ is used to accelerate the body, then $F^{'} = m a = 20 \times 3 = 60 N$
So, force of friction $= F - F^{'} = 80 - 60 = 20 N$

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A body of mass 20kg is kept initially at rest. A force of 80N is applied on the body then the acceleration produced in the body is 3m/s2, force of friction acting on the body is

The correct option is C 20NMass of the body, m=20kgApplied force, F=80NAcceleration produced, a=3m/s2If force F′ is used to accelerate the body, then F′=ma=20×3=60NSo, force of friction =F−F′=80−60=20N

A body of mass $20 k g$ is kept initially at rest. A force of $80 N$ is applied on the body then the acceleration produced in the body is $3 m / s^{2}$ , force of friction acting on the body is

The correct option is C $20 N$
Mass of the body, $m = 20 k g$
Applied force, $F = 80 N$
Acceleration produced, $a = 3 m / s^{2}$
If force $F^{'}$ is used to accelerate the body, then $F^{'} = m a = 20 \times 3 = 60 N$
So, force of friction $= F - F^{'} = 80 - 60 = 20 N$