A block of mass 2kg is lying on a floor. The coefficient of friction is 0.4. If a force of 2.5N is applied on the block as shown in the figure , then the magnitude of frictional force will be:
A
7.5N
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B
10N
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C
2.5N
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D
5N
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Solution
The correct option is C2.5N Given: μ=0.4F=2.5N m=2kg N=mg=20N The max value of friction =μN=0.4×20=8N Since, the force applied is 2.5N there is no need for friction to attain its max. value. ∴ Frictional force =2.5N