A block of mass 5 kg is kept against an acce | vedclass

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(B) For a block to remain steady on an accelerating wedge, the forces acting on the block in the frame of the wedge are: gravity ($mg$ downwards), pse

Vedclass · Accepted Answer

$\frac{30}{7} \ m \ s^{-2}$

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(B) For a block to remain steady on an accelerating wedge, the forces acting on the block in the frame of the wedge are: gravity ($mg$ downwards), pseudo force ($ma$ horizontally outwards), normal force ($N$ perpendicular to the surface), and friction ($f$ parallel to the surface).To find the minimum acceleration $(a_{\min})$, friction must act upwards along the incline to prevent the block from sliding down.Resolving forces parallel and perpendicular to the incline:$N = mg \cos 	heta + ma \sin 	heta$$f + ma \cos 	heta = mg \sin 	heta$Since $f = \mu N$, we have $\mu(mg \cos 	heta + ma \sin 	heta) = mg \sin 	heta - ma \cos 	heta$.Rearranging for $a$:$ma(\mu \sin 	heta + \cos 	heta) = mg(\sin 	heta - \mu \cos 	heta)$$a = g \frac{\sin 	heta - \mu \cos 	heta}{\cos 	heta + \mu \sin 	heta}$Given $	heta = 45^{\circ}$, $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.$a_{\min} = g \frac{\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \mu \frac{1}{\sqrt{2}}} = g \frac{1 - \mu}{1 + \mu}$Substituting $g = 10 \ m \ s^{-2}$ and $\mu = 0.4$:$a_{\min} = 10 	imes \frac{1 - 0.4}{1 + 0.4} = 10 	imes \frac{0.6}{1.4} = 10 	imes \frac{6}{14} = \frac{60}{14} = \frac{30}{7} \ m \ s^{-2}$.

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