Downside assertion (simplified)
An object slides down a ramp at angle θ to come across degree floor. Each surfaces have kinetic friction: μ’ on the ramp, μ on the extent. The thing reaches the bottom at velocity u. What’s its velocity when first totally on the extent?
(Unique is at https://www.physicsforums.com/threads/distanceablockslidesalongasurfacewithfrictiongivenwithaninitialvelocity.1047556/.)
There are a number of lacking particulars, which suggests the creator ignored sure subtleties that may come up in friction issues.
For concreteness, I’ll take the article to be a uniform rectangular block base size B, top A, diagonal 2r, and A/B=tan(α).
I’ll assume it’s meant that each ends of the bottom of the block ought to stay involved with the surfaces always.
Suppose the transition from the ramp to the extent floor is through an arc of radius R. The diagram provided implies R is vanishingly small, however I’ll begin with a less complicated case. In any occasion, I’ll assume u^{2}>>g max(R,r), which makes gravity irrelevant.
There are two fundamental instances to think about:

r<<R

r>>R
Case 1: r<<R
The utmost rotational KE might be small c.w. the translational KE and might be ignored.
Traversing the arc, the conventional power is the centripetal power, mu^{2}/R, the frictional power is μ mu^{2}/R, the trail size is Rθ, and the work performed in opposition to friction is μ mu^{2}θ.
The fraction of KE misplaced to friction is 2μ θ. E.g. with θ=π/4, for any μ>2/ π the article won’t attain the horizontal floor.
Within the unique downside, θ=π/6, μ=0.3 for the horizontal floor, 0.2 for the ramp. Utilizing the common, 0.25, the fraction misplaced is π/12, about 26%
Case 2: r>>R
We are able to deal with this case as an influence, i.e. R=0.
To simplify the algebra, all forces and impulses might be taken to be per unit mass.
An answer posted within the thread treats it merely as an inelastic influence within the vertical. Consequently, it finds v = u cos(θ), a lack of 25% of the KE. This overlooks frictional impulse.
There are three levels within the transition to horizontal movement:

The forefront strikes the bottom and begins to maneuver horizontally.

The block continues to slip, the forefront on the bottom, and the trailing edge on the ramp.

The trailing edge strikes the bottom.
There are frictional losses throughout every stage.
Stage 1
When the article strikes the bottom, there might be a vertical impulse J from the bottom at the forefront and an impulse J’ from the ramp, regular to the ramp, on the trailing fringe of the article.
Correspondingly, a frictional impulse μJ horizontally at the forefront and a frictional impulse μ’J’ from the ramp, up the ramp, on the trailing edge.
Let the horizontal velocity of the forefront instantly after influence be v and the rotation charge of the article simply after influence be ω. This means the horizontal velocity of the centre of the article simply after influence is vωr sin(θ+α), and its vertical velocity is ωr cos(θ+α) (down).
Horizontal momentum conservation:
μJ+ μ’J’ cos(θ) – J’ sin(θ) = u cos(θ) v + ωr sin(θ+α)
Vertical momentum conservation:
J+J’ cos(θ) + μ’J’ sin(θ) = u sin(θ) – ωr cos(θ+α)
Because the trailing edge continues parallel to the ramp,
v sin(θ) = ωB = 2ωr cos(α)
The second of inertia of the block is mr^{2}/3.
Angular momentum conservation about block centre:
ωr^{2}/3 = Jr cos(θ+α) – μJr sin(θ+α) – J’r cos(α) – μ’J’r sin(α)
cancelling r:
ωr/3 = J cos(θ+α) – μJ sin(θ+α) – J’ cos(α) – μ’J’ sin(α)
We now have 4 equations and 4 unknowns: J, J’, ω, v.
What to decide on for α? The unique diagram exhibits a sq. block, implying α= π/4, it specifies θ =π/6, and units μ=0.3 for the horizontal floor. Nevertheless, it seems that with these angles the again finish of the block would lose contact with the ramp when the forefront hits the bottom even with no friction.
We might compromise by utilizing α= 0.49, about 28°, which simply avoids lack of contact. This provides a lack of 37% of the KE, and that is simply the primary stage. (Even α= 0 provides a loss > 25%.)
As a substitute, I shall simplify issues by setting α= 0. That cuts the stage 1 loss to simply over 25%.
Stage 2
Regardless of selecting α= 0, the equations for this dynamic case are difficult.
At first, it might appear much like Case 1, however be aware the trajectory of the mass centre. Taking the centre of the rod object (as now assumed) to be at (x,y) relative to the underside of the ramp, (x+2y cot(θ))^{2}+y^{2}=r^{2}, an ellipse. I.e. it arcs downwards, requiring a downward centripetal power.
This means lack of contact at enough preliminary velocity.
Stage 3
Suppose, simply earlier than the trailing edge hits the bottom, the forefront is travelling at velocity v’ horizontally. The trailing edge will need to have been travelling at v’ horizontally additionally, and v’ tan(θ) vertically down. The mass centre of the block was subsequently descending at velocity v’ tan(θ)/2. On hitting the bottom full size, there’s a vertically upward impulse mv’ tan(θ)/2. and a frictional impulse μmv’ tan(θ)/2.
KE simply earlier than influence = mv’^{2}/2+mv’^{2}tan^{2}(θ)/8
KE simply after influence = mv’^{2}(1μ tan(θ)/2)^{2}/2
Fraction of preliminary KE remaining = (1 μ tan(θ)/2)^{2}/(1+tan^{2}(θ)/4)
With θ=π/6 and μ=0.3 that’s about 0.77, or a 23% loss.
Different shapes
How might the issue description be adjusted to make the evaluation easier?
A technique can be to specify the form of the article such that it doesn’t rotate. It could possibly be a triangle ABC, the place AC contacts the ramp and BC, the decrease edge, is horizontal. So ∠ABC = πθ.
With tan(∠ ACB)=tan(θ)/2, the mass centre can be vertically above B. This might tip neither whereas accelerating down the ramp nor when decelerating on the horizontal.
At influence with the bottom, there’s a vertically upward impulse mv sin(θ) and a corresponding horizontal frictional impulse μmv sin(θ). The ensuing horizontal velocity is subsequently v(cos(θ) – μ sin(θ)). With θ=π/6 and μ=0.3 that’s about 0.716v, a 49% lack of KE.
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