A demonstration apparatus on a table in the l | vedclass

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(B) The marble rolls without slipping. The force of static friction acts on the marble,but it does no work because the point of contact is instantaneo

Vedclass · Accepted Answer

$h_1 = h$,since total mechanical energy is conserved.

Vedclass · Answer

(B) The marble rolls without slipping. The force of static friction acts on the marble,but it does no work because the point of contact is instantaneously at rest relative to the track. Since we are neglecting air resistance and any other non-conservative forces (like rolling resistance),the total mechanical energy of the system is conserved. Initially,the marble is at rest at height $h$,so its total mechanical energy is $E_i = mgh$. At the highest point on the other side,the marble briefly stops at height $h_1$. Its velocity is zero,so its kinetic energy (both translational and rotational) is zero. Its total mechanical energy is $E_f = mgh_1$. By the law of conservation of mechanical energy,$E_i = E_f$,which implies $mgh = mgh_1$,or $h_1 = h$. Therefore,the height reached on the other side is equal to the initial height,regardless of the angles of inclination $	heta$ and $\phi$.

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