Q190 / 200 · TS EAMCET 2001 · Physics

A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5 , the acceleration is ( $g=$ Acceleration due to gravity $)$

  • A
    $\frac{g}{2}$
  • B
    $\frac{g}{2 \sqrt{2}}$
  • Correct answer
    C
    $\frac{3 g}{2 \sqrt{2}}$
  • D
    $\frac{g}{\sqrt{2}}$

Solution

Acceleration of an object up a rough inclined plane $\begin{aligned} a & =g(\sin \theta+\mu \cos \theta)=g\left(\sin 45^{\circ}+0.5 \cos 45^{\circ}\right) \\ & =g\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)=\frac{g}{\sqrt{2}}\left(1+\frac{1}{2}\right)=\frac{3 g}{2 \sqrt{2}}\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions