A particle is projected up along a rough incl | vedclass

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Question

(C) When a particle is projected up an inclined plane, both the component of gravity acting down the plane and the frictional force act in the directi

Vedclass · Accepted Answer

$\frac{3 g}{2 \sqrt{2}}$

Vedclass · Answer

(C) When a particle is projected up an inclined plane, both the component of gravity acting down the plane and the frictional force act in the direction opposite to the motion.The force acting down the plane is $F = mg \sin 	heta + f_k$, where $f_k = \mu N = \mu mg \cos 	heta$.Using Newton's second law, $ma = -(mg \sin 	heta + \mu mg \cos 	heta)$.The magnitude of the retardation (deceleration) is $a = g(\sin 	heta + \mu \cos 	heta)$.Given $	heta = 45^{\circ}$ and $\mu = 0.5 = \frac{1}{2}$.Substituting these values: $a = g(\sin 45^{\circ} + 0.5 \cos 45^{\circ})$.$a = g\left(\frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}}ight)$.$a = g\left(\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}ight) = g\left(\frac{2+1}{2\sqrt{2}}ight) = \frac{3g}{2\sqrt{2}}$.

$A$ 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).

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