COMEDK · Physics · 5. Laws of Motion
A block of mass \(0.1 \mathrm{~kg}\) is held against a wall by applying a horizontal force of \(5 \mathrm{~N}\) on it. If \(\mu_{s}\) between the wall and the block is \(0.5\), the magnitude of the frictional force acting on the block is
- A \(0.98 \mathrm{~N}\)
- B \(0.49 \mathrm{~N}\)
- C \(4.9 \mathrm{~N}\)
- D \(2.5 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(0.98 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Since, the block is at rest, the upward frictional force is equal to the weight or gravitational pull of the Earth i.e., \[ F_{S}=w=m g=0.1 \times 9.8=0.98 \mathrm{~N} \]
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