Q.Three identical solid spheres move down through three inclined planes A, B and C, all of the same dimensions. A is without friction, B is undergoing pure rolling, and C is rolling with slipping. Compare the kinetic energies $E_A$, $E_B$ and $E_C$ at the bottom.

Step 1. Set up the common starting point — all three spheres begin with the same gravitational PE.

Since all three inclines have identical dimensions and the spheres are identical, each sphere starts with the same gravitational potential energy $mgh$ (relative to the bottom of the incline) and each sphere's total mechanical energy budget available to convert during the descent is exactly $mgh$ in every case. The question is how much of that budget survives as kinetic energy at the bottom — and, in case C, how much is instead lost to heat.

Step 2. Analyse sphere A — frictionless incline, pure sliding, no rotation at all.

With NO friction anywhere along incline A, there is no tangential force at the contact point at all — so nothing ever applies a torque to spin the sphere up. The sphere slides down without rotating in the slightest. Since there is also no friction to dissipate energy as heat, mechanical energy is exactly conserved:

so $E_A=mgh$, entirely as translational kinetic energy.

Step 3. Analyse sphere B — pure rolling without slipping, static friction present but doing NO work.

Here static friction DOES act (it is in fact what causes the sphere to start rotating in the first place, supplying the torque about the center as derived from the rolling equations of §5.6.4), but because there is no relative SLIDING between the contact point and the incline surface in pure rolling (the point of contact is instantaneously at rest), static friction does zero work — it changes how the energy is DISTRIBUTED between translational and rotational forms, but it removes none of it. So mechanical energy is, once again, exactly conserved:

so $E_B=mgh$ as well — but now split, in the fixed ratio $1:K^2/R^2$ for a solid sphere ($K^2/R^2=2/5$), as $\tfrac57 mgh$ translational and $\tfrac27 mgh$ rotational.

Step 4. Analyse sphere C — rolling WITH slipping, kinetic friction present and dissipative.

On incline C, the sphere both translates and rotates, but not in the pure-rolling ratio ($v_{CM}\ne R\omega$) — there IS relative sliding between the contact point and the surface. Wherever there is relative sliding, the friction acting is KINETIC friction, and kinetic friction, acting over a nonzero relative sliding distance, does negative work on the system and converts some mechanical energy irreversibly into heat. So here the energy budget genuinely splits three ways:

meaning $KE_C = mgh - Q_{heat} < mgh$.

Step 5. Compare all three. …

Energy-conservation reasoning contrasting frictionless / no-slip …

• Assuming friction always dissipates energy, and therefore wrongly ranking B below A — static friction in PURE rolling does zero work because there is no relative sliding at the contact point, so B loses nothing, unlike C where slipping DOES cause dissipative kinetic friction. …