A point particle of mass m moves along the un | vedclass

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(A) The height of point $P$ is $h = 2 \ m$. The length of the inclined track $PQ$ is $L = h / \sin(30^\circ) = 2 / 0.5 = 4 \ m$.The energy lost over t

Vedclass · Accepted Answer

$0.29$ and $3.5 \ m$

Vedclass · Answer

(A) The height of point $P$ is $h = 2 \ m$. The length of the inclined track $PQ$ is $L = h / \sin(30^\circ) = 2 / 0.5 = 4 \ m$.The energy lost over the path $PQ$ is $W_{PQ} = \mu mg \cos(30^\circ) 	imes L = \mu mg (\sqrt{3}/2) 	imes 4 = 2\sqrt{3} \mu mg$.The energy lost over the horizontal path $QR$ is $W_{QR} = \mu mg x$.Given that the energy lost over $PQ$ and $QR$ are equal,$W_{PQ} = W_{QR}$:$2\sqrt{3} \mu mg = \mu mg x \implies x = 2\sqrt{3} \approx 3.46 \ m$.The total energy lost is equal to the initial potential energy of the particle: $W_{PQ} + W_{QR} = mgh$.Since $W_{PQ} = W_{QR}$,we have $2 W_{PQ} = mgh$,which means $W_{PQ} = mgh / 2$.$2\sqrt{3} \mu mg = mgh / 2 \implies 2\sqrt{3} \mu = h / 2$.Substituting $h = 2 \ m$: $2\sqrt{3} \mu = 1 \implies \mu = 1 / (2\sqrt{3}) \approx 1 / 3.464 \approx 0.288 \approx 0.29$.Thus,$\mu \approx 0.29$ and $x \approx 3.5 \ m$.

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