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Q. A particle is projected along a line of greatest slope on a rough plane inclined at an angle of $45^\circ $ with the horizontal. If the coefficient of friction is $\frac{1}{2}$ , then the retardation is

NTA AbhyasNTA Abhyas 2022

Solution:

The acceleration of the body on the incline is
$a=g\left(sin \theta + \mu cos ⁡ \theta \right)$
$a=g\left(sin 4 5 ^\circ + \frac{1}{2} \times cos ⁡ 45 ^\circ \right)$
$a=g\left(\frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}}\right)$
$a=\frac{g}{\sqrt{2}}\left(1 + \frac{1}{2}\right)$
$a=\frac{3 g}{2 \sqrt{2}}$