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(D) Let $m$ be the mass of the block and $	heta = 45^{\circ}$.The force required to push the block up the inclined plane is $F_1 = mg \sin 	heta + \

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let $m$ be the mass of the block and $	heta = 45^{\circ}$.The force required to push the block up the inclined plane is $F_1 = mg \sin 	heta + \mu mg \cos 	heta$.The force required to prevent the block from sliding down is $F_2 = mg \sin 	heta - \mu mg \cos 	heta$.Given that $F_1 = 3 F_2$,we have:$mg(\sin 45^{\circ} + \mu \cos 45^{\circ}) = 3 mg(\sin 45^{\circ} - \mu \cos 45^{\circ})$.Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we can cancel $mg$ and $\frac{1}{\sqrt{2}}$ from both sides:$1 + \mu = 3(1 - \mu)$.Expanding the equation:$1 + \mu = 3 - 3 \mu$.Rearranging the terms:$4 \mu = 2$.$\mu = 0.5$.Given $N = 10 \mu$,we calculate:$N = 10 	imes 0.5 = 5$.

$A$ block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10 \mu$,then $N$ is

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