The acceleration of a body of mass m sliding | vedclass

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(B) Consider a body of mass $m$ on an inclined plane with angle $	heta$. The forces acting on the body are:$1$. The component of gravitational force

Vedclass · Accepted Answer

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

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(B) Consider a body of mass $m$ on an inclined plane with angle $	heta$. The forces acting on the body are:$1$. The component of gravitational force acting down the plane: $mg \sin 	heta$.$2$. The normal force perpendicular to the plane: $N = mg \cos 	heta$.$3$. The kinetic frictional force acting up the plane: $f_k = \mu N = \mu mg \cos 	heta$.The net force $F_{	ext{net}}$ acting down the plane is the difference between the gravitational component and the frictional force:$F_{	ext{net}} = mg \sin 	heta - f_k = mg \sin 	heta - \mu mg \cos 	heta$.Using Newton's second law,$F_{	ext{net}} = ma$,we get:$ma = mg \sin 	heta - \mu mg \cos 	heta$.Dividing both sides by $m$,the acceleration $a$ is:$a = g(\sin 	heta - \mu \cos 	heta)$.

The acceleration of a body of mass $m$ sliding down an inclined plane with an angle of inclination $\theta$ and a coefficient of kinetic friction $\mu$ is:

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