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Question

(A) For a body of mass $m$ on an inclined plane with angle of inclination $	heta$ and coefficient of friction $\mu$,the forces acting along the plane

Vedclass · Accepted Answer

$g(\sin 	heta - \mu \cos 	heta)$

Vedclass · Answer

(A) For a body of mass $m$ on an inclined plane with angle of inclination $	heta$ and coefficient of friction $\mu$,the forces acting along the plane are the component of gravity $mg \sin 	heta$ acting downwards and the frictional force $f = \mu N$ acting upwards.The normal force $N$ is given by $N = mg \cos 	heta$.The net force $F_{	ext{net}}$ acting down the plane is $F_{	ext{net}} = mg \sin 	heta - f = mg \sin 	heta - \mu mg \cos 	heta$.Using Newton's second law,$F_{	ext{net}} = ma$,we get $ma = mg(\sin 	heta - \mu \cos 	heta)$.Therefore,the acceleration $a$ is $g(\sin 	heta - \mu \cos 	heta)$.

When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal,its acceleration is

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