Why is the velocity at the cusp point equal to the final velocity on the hill?

[ButteredCatParadox]

Homework Statement

A sled goes down a frictionless hill onto flat ground with friction. The sled comes to a stop in some distance. The height of the hill, the angle of the hill w.r.t the horizontal, the initial velocity of the sled, and the coefficient of kinetic friction are all known. I want to solve this without using conservation of energy.

Homework Equations

Constant acceleration equations of motion, Newton's 2nd law.

The Attempt at a Solution

The velocity at the bottom of the hill can be found using the appropriate kinematics equation. This velocity can then be used as the initial velocity in solving how far the sled travels on the flat ground.

My question is regarding the velocity at the cusp point between hill and flat ground. What is a good argument for why the entire final velocity on the hill should be used for the initial velocity on the flat (as opposed the horizontal component)? If I draw a free body diagram at the cusp point, I want the forces to net to zero in the direction perpendicular to half the hill angle in order for direction, but not magnitude of velocity to change. When I do this, I get an expression that's not clear to me why it equals zero:
W*sin(Θ/2)-μ_k*N₂*cos(Θ/2)+N₁sin(Θ/2)-N₂sin(Θ/2)=0
where W=weight of sled, Θ=hill angle, μ_k=coefficient of kinetic friction, N₂=normal force of flat on sled, N₁=normal force of hill on sled. The values of N₁ and N₂ at the cusp point seem like they should be different than when the sled is either only on the hill or only on the flat.

I want to understand this without invoking conservation of energy.

 

[lychette]

Homework Statement

A sled goes down a frictionless hill onto flat ground with friction. The sled comes to a stop in some distance. The height of the hill, the angle of the hill w.r.t the horizontal, the initial velocity of the sled, and the coefficient of kinetic friction are all known. I want to solve this without using conservation of energy.

Homework Equations

Constant acceleration equations of motion, Newton's 2nd law.

The Attempt at a Solution

The velocity at the bottom of the hill can be found using the appropriate kinematics equation. This velocity can then be used as the initial velocity in solving how far the sled travels on the flat ground.

My question is regarding the velocity at the cusp point between hill and flat ground. What is a good argument for why the entire final velocity on the hill should be used for the initial velocity on the flat (as opposed the horizontal component)? If I draw a free body diagram at the cusp point, I want the forces to net to zero in the direction perpendicular to half the hill angle in order for direction, but not magnitude of velocity to change. When I do this, I get an expression that's not clear to me why it equals zero:
W*sin(Θ/2)-μ_k*N₂*cos(Θ/2)+N₁sin(Θ/2)-N₂sin(Θ/2)=0
where W=weight of sled, Θ=hill angle, μ_k=coefficient of kinetic friction, N₂=normal force of flat on sled, N₁=normal force of hill on sled. The values of N₁ and N₂ at the cusp point seem like they should be different than when the sled is either only on the hill or only on the flat.

I want to understand this without invoking conservation of energy.

Your 'appropriate kinematics equations ' are based on conservation of energy. Have you considered this?

 

[ButteredCatParadox]

Your 'appropriate kinematics equations ' are based on conservation of energy. Have you considered this?

Hi Lychette,

Thanks for the response.

I'm not sure I follow. The equation to which I was referring was:
v_f²=v_i²+2aΔx
which is derived just from the assumption that acceleration is constant.

This problem comes from a section before conservation of energy is introduced so it should be understandable w/o invoking those concepts.

 

[haruspex]

What is a good argument for why the entire final velocity on the hill should be used for the initial velocity on the flat (as opposed the horizontal component)?

There isn't one. It depends on details.
If there is a smooth transition from slope to horizontal, still without friction, then the whole velocity is transferred.
If it is very abrupt, there is effectively an impact in the vertical direction. Even if it is partly elastic it doesn't help because the vertical component will remain vertical, i.e. bouncing, and not contribute to the forward motion on the horizontal.
If it is a smooth curve but friction starts before the end of the curve then the normal force is a little stronger than on the level since it has to provide centripetal acceleration. Thus the frictional loss is a bit greater than later on.

 

[lychette]

Hi Lychette,

Thanks for the response.

I'm not sure I follow. The equation to which I was referring was:
v_f²=v_i²+2aΔx
which is derived just from the assumption that acceleration is constant.

This problem comes from a section before conservation of energy is introduced so it should be understandable w/o invoking those concepts.

The equation you are referring to is an indication of conservation of energy i.e. ...increase in KE = work done

 

[haruspex]

The equation you are referring to is an indication of conservation of energy i.e. ...increase in KE = work done

BCP is correct. Δ(v²) = 2as is a simple matter of kinematics.
It is equivalent to saying that the work done on the object equals its gain in KE, but that is always true. It is not affected by friction etc. Conservation of work is to do with whether this also matches the work done by other objects, gravity etc.

 

[ButteredCatParadox]

There isn't one. It depends on details.
If there is a smooth transition from slope to horizontal, still without friction, then the whole velocity is transferred.
If it is very abrupt, there is effectively an impact in the vertical direction. Even if it is partly elastic it doesn't help because the vertical component will remain vertical, i.e. bouncing, and not contribute to the forward motion on the horizontal.
If it is a smooth curve but friction starts before the end of the curve then the normal force is a little stronger than on the level since it has to provide centripetal acceleration. Thus the frictional loss is a bit greater than later on.

Thanks for the helpful response, haruspex.

How can one differentiate between a smooth and abrupt transition? If θ is 25°, could the instant of transition be treated as (uniform) circular motion?

 

[PeroK]

What is a good argument for why the entire final velocity on the hill should be used for the initial velocity on the flat (as opposed the horizontal component)?

One argument is that that is what you would practically expect in most circumstances. E.g. ski-jumper, car or bicycle getting to the bottom of a slope. They don't suddenly lose their vertical speed when they hit the flat. If I go downhill on my bike, I expect to take all that speed forward onto the flat. It would be pretty weird suddenly to slow down at the bottom of a hill!

 

[ButteredCatParadox]

One argument is that that is what you would practically expect in most circumstances. E.g. ski-jumper, car or bicycle getting to the bottom of a slope. They don't suddenly lose their vertical speed when they hit the flat. If I go downhill on my bike, I expect to take all that speed forward onto the flat. It would be pretty weird suddenly to slow down at the bottom of a hill!

Thanks for example, PeroK.

If there were an abrupt transition between hill and flat, wouldn't you expect to lose at least a little speed on your bike due to the impact?

 

[PeroK]

Thanks for example, PeroK.

If there were an abrupt transition between hill and flat, wouldn't you expect to lose at least a little speed on your bike due to the impact?

If it was too abrupt I might expect to fall off!

 

[lychette]

BCP is correct. Δ(v²) = 2as is a simple matter of kinematics.
It is equivalent to saying that the work done on the object equals its gain in KE, but that is always true. It is not affected by friction etc. Conservation of work is to do with whether this also matches the work done by other objects, gravity etc.

The simple matter of kinematics comes directly from analysing conservation of energy.
Gravitational PE can be converted to KE...Wh = (mv²)/2 or mgh = (mv²)/2 gives v² = gh
Also KE = force x displacement... (mv²)/2 = F x s = mas giving v² = 2as
BCP wants to avoid using conservation of energy but by using the kinematic equations he is effectively using conservation of energy.
It would be a strange world if the kinematic equations were in some way separate from conservation of energy.