ExamBro
ExamBro
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 \(=10 \mathrm{~ms}^{-2}\) )

  1. A 0.5 N
  2. B 2 N
  3. C 4 N
  4. D 8 N
Verified Solution

Answer & Solution

Correct Answer

(C) 4 N

Step-by-step Solution

Detailed explanation

\(F \cos 45^{\circ} = \mu_s N\) \(N = mg - F \sin 45^{\circ}\) \(F \cos 45^{\circ} = \mu_s (mg - F \sin 45^{\circ})\) \(F \cdot \frac{1}{\sqrt{2}} = 0.25 \left( \sqrt{2} \cdot 10 - F \cdot \frac{1}{\sqrt{2}} \right)\) \(\frac{F}{\sqrt{2}} = 2.5\sqrt{2} - \frac{0.25F}{\sqrt{2}}\)…