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A block of mass 4 kg is placed on a floor. The coefficient of friction between block and floor is 0.2. If a 6N force is applied on the block, what is the value of acceleration of block? (Take g = 10m/s2 ) please solve
Asked by aryanbalwan | 11 Jan, 2023, 06:31: PM
Given that,
Mass = 4 kg
Coefficient of static friction = 0.2
Force = 6 N
Now,
The external force needed to overcome the static friction is given as
Fs​ =μmg = 0.2x 4x10 = 8 N
Hence for the given case the applied force is not enough to move the block, thus the acceleration will be zero.
i.e., a = 0 (Applied force < Limiting value of friction)
Answered by Jayesh Sah | 12 Jan, 2023, 07:09: PM

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