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(A) When the block moves up the inclined plane,the forces acting against its motion are the component of gravitational force along the plane $(mg \sin

Vedclass · Accepted Answer

$\frac{15}{\sqrt{2}}$

Vedclass · Answer

(A) When the block moves up the inclined plane,the forces acting against its motion are the component of gravitational force along the plane $(mg \sin 	heta)$ and the kinetic frictional force $(f_k = \mu N = \mu mg \cos 	heta)$.The total retarding force $F_{ret} = mg \sin 	heta + \mu mg \cos 	heta$.Using Newton's second law,the retardation $a$ is given by $a = \frac{F_{ret}}{m} = g \sin 	heta + \mu g \cos 	heta$.Given $g = 10 \, ms^{-2}$,$	heta = 45^{\circ}$,and $\mu = 0.5$:$a = 10 \sin 45^{\circ} + 0.5 	imes 10 \cos 45^{\circ}$$a = 10 \left( \frac{1}{\sqrt{2}} ight) + 5 \left( \frac{1}{\sqrt{2}} ight)$$a = \frac{10 + 5}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, ms^{-2}$.

$A$ block of mass $1 \, kg$ is projected from the lowest point up along an inclined plane with an angle of inclination $45^{\circ}$ and coefficient of friction $\mu = 0.5$. If $g = 10 \, ms^{-2}$,the retardation experienced by the block is ............. $ms^{-2}$.

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