AP EAMCET · PHYSICS · Laws of Motion
A block of mass \(2 \mathrm{~kg}\) is being pushed against a wall by a force \(F=90 \mathrm{~N}\) as shown in the figure. If the coefficient of friction is 0.25 , then the magnitude of acceleration of the block is (Take, \(\left.g=10 \mathrm{~ms}^{-2}\right)\left(\sin 37^{\circ}=\frac{3}{5}\right)\)

- A \(16 \mathrm{~ms}^{-2}\)
- B \(8 \mathrm{~ms}^{-2}\)
- C \(38 \mathrm{~ms}^{-2}\)
- D \(54 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(B) \(8 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Block's weight (downward), \(w=2 \times 10=20 \mathrm{~N}\) Vertical component of applied force (upwards), \[ F_V=F \sin \theta=90 \times 3 / 5=54 \mathrm{~N} \] Maximum frictional force,…
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