TS EAMCET · Physics · Laws of Motion
A block of mass \(\sqrt{2} \mathrm{~kg}\) is placed on a rough horizontal surface. A force ' \(F\) ' acting upwards at an angle of \(45^{\circ}\) with the horizontal causes the block to start motion. If the coefficient of static friction between the surface and the block is 0.25, the magnitude of the force ' \(F\) ' is
(Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A 0.5 N
- B 2 N
- C 4 N
- D 8 N
Answer & Solution
Correct Answer
(C) 4 N
Step-by-step Solution
Detailed explanation
\(F \cos 45^{\circ} = \mu_s N\) \(N = mg - F \sin 45^{\circ}\) \(F \cos 45^{\circ} = \mu_s (mg - F \sin 45^{\circ})\) \(F \cdot \frac{1}{\sqrt{2}} = 0.25 \left( \sqrt{2} \cdot 10 - F \cdot \frac{1}{\sqrt{2}} \right)\) \(\frac{F}{\sqrt{2}} = 2.5\sqrt{2} - \frac{0.25F}{\sqrt{2}}\)…
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