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AIIMS 2014
Exam
AIIMS
Year
2014
Subject
Physics
Chapter
Laws of Motion

AIIMS 2014 · Physics · Laws of Motion

A block of mass 2 ~kg is placed on the floor. The coefficient of...

A block of mass 2 kg2 \mathrm{~kg} is placed on the floor. The coefficient of static friction is 0.4 . If a force of 2.8 N2.8 \mathrm{~N} is applied on the block parallel to the floor, the force of friction between the block and the floor is (g=10 m s−2)\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)

Options

  1. A2.8 N2.8 \mathrm{~N}
  2. B2 N2 \mathrm{~N}
  3. C8 N8 \mathrm{~N}
  4. DZero

Correct answer

A. 2.8 N2.8 \mathrm{~N}

Step-by-step solution

 As fms=μsmg=0.4×2×10=8 N. \text{ As } f_{m s}=\mu_s m g=0.4 \times 2 \times 10=8 \mathrm{~N} \text{. } The applied force of 2.8 N2.8 \mathrm{~N} is less than fms(=8 N)f_{m s}(=8 \mathrm{~N}) and as such the block does not move. When the block is at rest, frictional force = applied force =2.8 N=2.8 \mathrm{~N}
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