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  4. A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor (Taking g=10 ms -2 ) is

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Q. A block of mass $2 \,kg$ is placed on the floor. The coefficient of static friction is $0.4$. If a force of $2.8 \,N$ is applied on the block parallel to floor, the force of friction between the block and floor
(Taking $g=10\, ms ^{-2}$ ) is

Laws of Motion

Solution:

Minimum force required to move the block
$=\mu R=\mu\, mg =0.4 \times 2 \times 10=8\, N$
Since the force applied is only $2.8\, N$, the block fails to move and static fraction $=$ applied force $=2.8\, N$