How this calculator works
A sliding block loses mechanical energy to kinetic friction. A rolling body instead divides energy between translation and rotation; ideal static friction does no work. The rolling model assumes the initial spin already matches the initial downhill speed and checks whether the supplied static coefficient can prevent slipping. The torus rolls upright on its outer rim and uses its symmetry-axis inertia. Air drag and rolling deformation are excluded.
Formula
L=H/sin(theta); a_slide=g(sin(theta)-mu cos(theta)); a_roll=g sin(theta)/(1+I/(mr^2)); v^2=u^2+2aL; t=2L/(u+v).
Worked example
For a 0.5 kg sliding block on a slope with sin(theta)=0.6, cos(theta)=0.8, mu=0.25 and g=10 m/s2, acceleration is 4 m/s2. An 8 m run from rest takes 2 s and ends at 8 m/s, losing 8 J.
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Frequently asked questions
Why does a sphere accelerate more slowly than a frictionless block?
Some gravitational energy becomes rotational kinetic energy instead of centre-of-mass motion.
Is rolling friction the same as static friction?
No. This ideal rolling model uses static friction to enforce no slip. Deformation losses belong in a rolling-resistance model.
Can a block slow down on the incline?
Yes, if friction exceeds the downhill gravity component. It reaches the bottom only if its initial speed is sufficient.