Find the speed of the biker at the bottom of the hill

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In summary: Or using the work-energy equation...$$(F_I{_I} + F_F) d cos\theta = \frac{1}{2} m\cdot v_f^2$$Or using the principle of conservation of energy...$$ W = mv_f$$In summary, the biker will lose potential energy due to gravity and gain kinetic energy when rolling down an incline.
  • #1
paulimerci
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Homework Statement
A biker and his bike have a combined mass of 90kg. The rider starts from rest and rolls(no pedalling) down an incline that is 500m long. The plane is at an incline of 5 degrees above the horizontal. The average frictional force acting on the bike/rider is equal to 60N. Find the speed of the biker at the bottom of the hill.
Relevant Equations
Net work done on a system = change in K.E
I used the work-energy equation to solve this problem.Only the parallel component of force acts to accelerate the biker and increases K.E., so only the parallel component and frictional force does work on the biker.
therefore,
Net work done on a system = change in K.E
$$(F_I{_I} + F_F) d cos\theta = \frac{1}{2} m\cdot v_f^2$$
$$ (mg sin\theta \cdot d cos\theta) + (F_F d cos\theta) = \frac{1}{2} m\cdot v_f^2$$
$$ (90 \cdot 10 \cdot sin 5 \cdot 500 \cdot cos0) + (60 \cdot 500 \cdot cos 180) = \frac{1}{2} m\cdot v_f^2$$
$$ v_f = 14.31 \frac{m}{s}$$
where, ##cos 0## - angle between the force and the displacement, where the parallel component of force is in the direction of motion and ##cos180## - angle between the frictional force and the displacement, ##v_i =0##

I know something is not right in the above equations that I solved. I didn't include GPE which should give me a different set of equations, and I didn’t use that approach; I'm just confused. Any help would be greatly appreciated.
 

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  • #2
Use \parallel to get ## F_\parallel ##
Code:
## F_\parallel ##

Also use \cos to write ##\cos## function.
Code:
## \cos \theta ##

Now the frictional force is already along the motion, you should not include any angles there.

Why are you even using ## \cos \theta ##? You should view this is a one-dimensional problem. You already have the lenght of the incline which is the distance travelled by the bike.
 
  • #3
malawi_glenn said:
Now the frictional force is already along the motion
Isn't friction between the ground and a rolling object like the wheels of a bicycle acting in the direction of motion of the center of mass of the rolling object? Else the rolling object would slip in the absence of ground friction.
 
  • #4
vcsharp2003 said:
Isn't friction between the ground and a rolling object like the wheels of a bicycle acting in the direction of motion of the center of mass of the rolling object? Else the rolling object would slip in the absence of ground friction.
It could be the case, but since we are not given any information about the wheels (radius, moment of inertia etc) we will assume that the bike can be treated as a point-object.
 
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  • #5
malawi_glenn said:
It could be the case, but since we are not given any information about the wheels (radius, moment of inertia etc) we will assume that the bike can be treated as a point-object.
I see.
That can cause confusion. Taking different directions of frictional force based on given information.

By point mass, do you mean the motion of the center of mass of the bicycle system?
 
  • #6
malawi_glenn said:
It could be the case, but since we are not given any information about the wheels (radius, moment of inertia etc) we will assume that the bike can be treated as a point-object.
Friction between the bicycle system and air would always be opposite to the motion of center of mass. But, probably the air friction is too low to be considered in such motion problems.
 
  • #7
malawi_glenn said:
Use \parallel to get ## F_\parallel ##
Code:
## F_\parallel ##

Also use \cos to write ##\cos## function.
Code:
## \cos \theta ##

Now the frictional force is already along the motion, you should not include any angles there.

Why are you even using ## \cos \theta ##? You should view this is a one-dimensional problem. You already have the lenght of the incline which is the distance travelled by the bike.
Yeah, you're right. But still, do I get the same answer? I don't think so
 
  • #8
vcsharp2003 said:
Friction between the bicycle system and air would always be opposite to the motion of center of mass. But, probably the air friction is too low to be considered in such motion problems.
I meant along but in exact opposite direction
 
  • #9
You can solve it this way

Object will loose potential energy due to gravity.
This loss of energy will become friction work and gain in kinetic energy. I got 13.7 m/s
 
  • #10
paulimerci said:
Homework Statement:: A biker and his bike have a combined mass of 90kg. The rider starts from rest and rolls(no pedalling) down an incline that is 500m long. The plane is at an incline of 5 degrees above the horizontal. The average frictional force acting on the bike/rider is equal to 60N.
It's funny how many question setters think friction acts on a rolling wheel! It should be rolling resistance. The main thing, however, that would slow a bicycle going downhill would be air resistance.
paulimerci said:
$$ v_f = 14.31 \frac{m}{s}$$
This is not quite right, as you have not treated the slope correctly. If a slope is ##500 \ m## long, then it's ##500 \ m## long. It's not ##500 \cos \theta \ m## long.
paulimerci said:
I know something is not right in the above equations that I solved. I didn't include GPE which should give me a different set of equations, and I didn’t use that approach; I'm just confused. Any help would be greatly appreciated.
There are several ways to solve this problem. Using GPE is one, but not the only one.
 
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  • #11
PeroK said:
It's funny how many question setters think friction acts on a rolling wheel! It should be rolling resistance. The main thing, however, that would slow a bicycle going downhill would be air resistance.

This is not quite right, as you have not treated the slope correctly. If a slope is ##500 \ m## long, then it's ##500 \ m## long. It's not ##500 \cos \theta \ m## long.

There are several ways to solve this problem. Using GPE is one, but not the only one.
$$(F_\parallel - F_F) d = \frac{1}{2} m\cdot v_f^2$$
$$ (mg sin\theta \cdot d ) - (F_F d) = \frac{1}{2} m\cdot v_f^2$$
$$ (90 \cdot 10 \cdot sin 5 \cdot 500 ) - (60 \cdot 500) = \frac{1}{2} m\cdot v_f^2$$
$$ v_f = 14.31 \frac{m}{s}$$
I'm confused about whether the force of friction is negative in the above equation, because the net work done is adding the work done by the individual forces, and since it's a scalar quantity, do sign conventions matter here?
 
  • #12
paulimerci said:
. . . and since it's a scalar quantity, do sign conventions matter here?
Are you suggesting that scalars can only be positive? There are no sign conventions in this particular case. The work done by a force on an object is (as you know) $$W=F~d~\cos\!\theta.$$ Here
##F## = the magnitude of the force vector, a positive number.
##d## = the magnitude of the displacement vector, a positive number.
##\cos\!\theta## = the cosine of the angle between the two vectors. It is positive when the angle is between zero and 90°, negative when the angle is between 90° and 180° and zero when it is equal to 90°.

So the work (which is a scalar quantity) done by the force can be positive, negative or zero depending on the relative angle between the two vectors. There is no sign convention here and the result is independent of how you choose the coordinate system in which you describe the two vectors.
 
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  • #13
paulimerci said:
I'm confused about whether the force of friction is negative in the above equation, because the net work done is adding the work done by the individual forces, and since it's a scalar quantity, do sign conventions matter here?
Well, putting a ##+## sign in front of the friction term adds to total available energy to be converted to kinetic energy. Does that sound correct to you?
 
  • #14
kuruman said:
Are you suggesting that scalars can only be positive? There are no sign conventions in this particular case. The work done by a force on an object is (as you know) $$W=F~d~\cos\!\theta.$$ Here
##F## = the magnitude of the force vector, a positive number.
##d## = the magnitude of the displacement vector, a positive number.
##\cos\!\theta## = the cosine of the angle between the two vectors. It is positive when the angle is between zero and 90°, negative when the angle is between 90° and 180° and zero when it is equal to 90°.

So the work (which is a scalar quantity) done by the force can be positive, negative or zero depending on the relative angle between the two vectors. There is no sign convention here and the result is independent of how you choose the coordinate system in which you describe the two vectors
Thank you, that makes sense! I can understand that the sign of W is entirely determined by the angle ##\theta## between the force and the displacement vectors.
 
  • #15
I do admit that this does get confusing when we see COE as:

$$U_o + KE_o + PE_o = U_f + KE_f+PE_f + W_{nc}$$

Making ##W_{nc}## negative on the RHS is equivalent to making it positive on the LHS. Then when you have a force that is adding energy into the system its going to be negative on the RHS. Its seems kind of strange.
 
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  • #16
paulimerci said:
$$ v_f = 14.31 \frac{m}{s}$$
Okay, but if you use ##g = 10 \ m/s^2##, then you need to think about the number of significant figures in your answer.
paulimerci said:
I'm confused about whether the force of friction is negative in the above equation, because the net work done is adding the work done by the individual forces, and since it's a scalar quantity, do sign conventions matter here?
Friction (rolling resistance) must be up the slope, acting against motion down the slope.
 
  • #17
PeroK said:
Okay, but if you use ##g = 10 \ m/s^2##, then you need to think about the number of significant figures in your answer.

Friction (rolling resistance) must be up the

PeroK said:
slope, acting against motion down the slope.
Okay, so the answer would be ##v_f = 14 \ m/s
##
 
  • #18
erobz said:
I do admit that this does get confusing when we see COE as:

$$U_o + KE_o + PE_o = U_f + KE_f+PE_f + W_{nc}$$

Making ##W_{nc}## negative on the RHS is equivalent to making it positive on the LHS. Then when you have a force that is adding energy into the system its going to be negative on the RHS. Its seems kind of strange.
If we use COE, do we get the same answer?
 
  • #19
paulimerci said:
The average frictional force acting on the bike/rider is equal to 60N.
The problem setter has made a mistake here.

The intent is pretty clearly that this "average force" should be multiplied by total displacement to get the total work drained away by friction. But for that computation to work, the average must be a displacement-weighted average. A time-weighted average (the default sort of average if nothing else is specified) will not do the job.

If what you have in hand is a time-weighted average then you can multiply it by elapsed time to get impulse.

If what you have in hand is a displacement-weighted average then you can multiply it by total displacement to get work.

A time-weighted average will not help us since we do not know elapsed time and do not have sufficient information to compute it.
 
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  • #20
paulimerci said:
If we use COE, do we get the same answer?
You used that equation you quoted (that I wrote down). Thats conservation of energy, that allows for waste heat or added work. A precursor to the first law I thermodynamics that focuses on mechanical energy transformation. ##U## was a term that was supposed to encapsulate other forms of mechanical energy (like springs, etc.., not internal energy like it signifies in the First Law).

I'm not sure if it has a special name that is specific to the addition of the ##W_{nc}## term. I always refer to it as COE.
 
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  • #21
erobz said:
You used that equation you quoted (that I wrote down). Thats conservation of energy, that allows for waste heat or added work. A precursor to the first law I thermodynamics that focuses on mechanical energy transformation. ##U## was a term that was supposed to encapsulate other forms of mechanical energy (like springs, etc.., not internal energy like it signifies in the First Law).

I'm not sure if it has a special name that is specific to the addition of the ##W_{nc}## term. I always refer to it as COE.
At the top of the hill, the biker has a maximum GPE, and when there is no friction involved, GPE is transformed into K.E. But in this case frictional force is doing negative work which takes away some of the GPE. Using the law of conservation of energy,
Total energy initial = Total energy final
$$ mgh = \frac{1}{2}mv^2 + W_F $$
$$mgh - F_F \cdot d \cos \theta = \frac {1}{2}mv_f^2$$
##\theta = 180## which will add the energies on the LHS of the equation, which should not be the case.
 

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  • #22
paulimerci said:
At the top of the hill, the biker has a maximum GPE, and when there is no friction involved, GPE is transformed into K.E. But in this case frictional force is doing negative work which takes away some of the GPE. Using the law of conservation of energy,
Total energy initial = Total energy final
$$ mgh = \frac{1}{2}mv^2 + W_F $$
$$mgh - F_F \cdot d \cos \theta = \frac {1}{2}mv_f^2$$
##\theta = 180## which will add the energies on the LHS of the equation, which should not be the case.
When using that you have to apply some interpretation and intuition.

For this "mechanical energy robbing" process, the LHS is ALL the energy you have available. The RHS represents all the bins that energy gets split up into. So indeed, it is not consistent with the strict\formal definition of work as it appeared above.

I don't know why I don't see this in my intro physics textbook, but maybe all we need to be consistent with the definition of work would be to change the sign to a negative on the original equation

$$ U_o + KE_o + PE_o = U_f + KE_f + PE_f - \int \mathbf{\vec{F_{nc}}} \cdot d \mathscr{ \vec{ l }}$$

Like I said, I don't apply it like that, I always just "think about it" for a given situation and hopefully apply appropriate sign to the ##W_{nc}## term. The textbook actually suggests that approach too. Maybe a minus sign on that last term runs into issues somewhere else I'm not seeing currently, maybe they thought it more simple to remember with all ##+##'s in the equation. The Physicist's will probably speak to this better.
 
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  • #23
I think one needs to be clear what system one is using here and whether that system is open or closed. Invoking mechanical energy conservation is inappropriate because ##K+U## is not conserved since there is friction. That leaves (I) the work energy theorem and (II) total energy conservation.

Case I
Assume that the system is the cyclist and bicycle. There are two entities doing external work on this system, the Earth and the whatever provides this "average" force of friction. Then the work energy theorem says $$\begin{align} & \Delta K=W_{\!g}+W_{\!f} \nonumber \\ &\frac{1}{2}mv_{\!f}^2-0=m~g~d\sin\!\theta-F_F ~d.\nonumber \end{align}$$Case II
Assume that the system is the cyclist + bicycle + Earth and that this is an isolated system, i.e. no external entities do work on it. Conservation of total energy is expressed as ##\Delta E_{\text{tot}}=0.## We have to account for the energy transformations within the system. These are the changes in kinetic and potential energy as well as changes in thermal energy due to friction. Specifically, the work done by friction ##W_{\!f}## appears as an increase of the thermal energy of the system, cyclist and bicycle warm up and the air molecules around them moves faster on average. We write ##\Delta E_{\text{therm}}=-W_{\!f}=F_F~d.## Then by the principle of total energy conservation, $$0=\Delta E_{\text{tot}}=\Delta K + \Delta U_{\!g}+\Delta E_{\text{therm}}=\left(\frac{1}{2}mv_{\!f}^2 -0\right)+(0-m~g~d\sin\!\theta)+F_F~d.$$The equations say the same thing, but the terms in them are shifted around because the energy accounting is different.
 
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  • #24
kuruman said:
I think one needs to be clear what system one is using here and whether that system is open or closed. Invoking mechanical energy conservation is inappropriate because ##K+U## is not conserved since there is friction. That leaves (I) the work energy theorem and (II) total energy conservation.

Case I
Assume that the system is the cyclist and bicycle. There are two entities doing external work on this system, the Earth and the whatever provides this "average" force of friction. Then the work energy theorem says $$\begin{align} & \Delta K=W_{\!g}+W_{\!f} \nonumber \\ &\frac{1}{2}mv_{\!f}^2-0=m~g~d\sin\!\theta-F_F ~d.\nonumber \end{align}$$Case II
Assume that the system is the cyclist + bicycle + Earth and that this is an isolated system, i.e. no external entities do work on it. Conservation of total energy is expressed as ##\Delta E_{\text{tot}}=0.## We have to account for the energy transformations within the system. These are the changes in kinetic and potential energy as well as changes in thermal energy due to friction. Specifically, the work done by friction ##W_{\!f}## appears as an increase of the thermal energy of the system, cyclist and bicycle warm up and the air molecules around them moves faster on average. We write ##\Delta E_{\text{therm}}=-W_{\!f}=F_F~d.## Then by the principle of total energy conservation, $$0=\Delta E_{\text{tot}}=\Delta K + \Delta U_{\!g}+\Delta E_{\text{therm}}=\left(\frac{1}{2}mv_{\!f}^2 -0\right)+(0-m~g~d\sin\!\theta)+F_F~d.$$The equations say the same thing, but the terms in them are shifted around because the energy accounting is different.
Thank you for the detailed explanation. In in the above statement
##\Delta E_{\text{therm}}=-W_{\!f}=F_F~d.## why it's not ##-W_{\!f} = -F_F~d##? Did you take the angle between the frictional force vector and the displacement vector as ##180 ##?
 
  • #25
Note that the attributions in #24 immediately above were broken. I believe that I have unravelled things correctly here.
paulimerci said:
kuruman said:
##\Delta E_{\text{therm}}=-W_{\!f}=F_F~d.##
Why is it not ##-W_f = - F_F d##

Did you take the angle between the frictional force vector and the displacement vector as 180°
What, exactly, is ##F_F## in your view? In my view, it is an unsigned scalar specified as being 60 N. It is the magnitude of the frictional force between bicycle and ramp.

We need not concern ourselves over much with the direction of either force or displacement. We should intuitively realize that the net work done by the frictional force between two objects within a single system will always be negative. For the sorts of surfaces one encounters in first year physics, the force will always be anti-parallel to the slip direction This means that we can be quite casual about sign conventions. We can multiply the magnitude of the force pair by the magnitude of the displacement, negate the result and call it a day:$$W_f = - F_F\ d$$
Clearly, thermal energy will be created by the friction. The change will be positive. If we are multiplying +60 N by +500 m we do not want to obtain a negative result.

[I trust the idea that my hands get warm when I rub them together a lot more than I trust myself to get all of the sign conventions right when deduce the same result from a vector equation].
 
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  • #26
jbriggs444 said:
Note that the attributions in #24 immediately above were broken. I believe that I have unravelled things correctly here.

What, exactly, is ##F_F## in your view? In my view, it is an unsigned scalar specified as being 60 N. It is the magnitude of the frictional force between bicycle and ramp.

We need not concern ourselves over much with the direction of either force or displacement. We should intuitively realize that the net work done by the frictional force between two objects within a single system will always be negative. For the sorts of surfaces one encounters in first year physics, the force will always be anti-parallel to the slip direction This means that we can be quite casual about sign conventions. We can multiply the magnitude of the force pair by the magnitude of the displacement, negate the result and call it a day:$$W_f = - F_F\ d$$
Clearly, thermal energy will be created by the friction. The change will be positive. If we are multiplying +60 N by +500 m we do not want to obtain a negative result.

[I trust the idea that my hands get warm when I rub them together a lot more than I trust myself to get all of the sign conventions right when deduce the same result from a vector equation].
Thank you, I hope I understand, but at the same time, I'm not confident enough when it comes to sign conventions. Thank you for taking the time to explain in detail. Thanks to everyone!
 
  • #27
jbriggs444 said:
What, exactly, is ##F_F## in your view? In my view, it is an unsigned scalar specified as being 60 N. It is the magnitude of the frictional force between bicycle and ramp.
That coincides with my view.
jbriggs444 said:
We need not concern ourselves over much with the direction of either force or displacement. We should intuitively realize that the net work done by the frictional force between two objects within a single system will always be negative.
Precisely. That is why I think that total energy conservation (Case II, post #23) of an isolated multicomponent system is a better way to approach problems of this sort. Then one can write an equation in which the change in total energy is $$\Delta E_{\text{tot}}=\Delta K + \Delta U_{\!g}+\Delta E_{\text{therm}}=0$$ in which ##\Delta E_{\text{therm}} > 0##. That is always the case. Then one has to find a recipe for the change in thermal energy. For a block sliding on the surface, the recipe says ##\Delta E_{\text{therm}} =\mu_kmgd## where all quantities are positive. For current running through a resistor the recipe says ##\Delta E_{\text{therm}} =I^2R\Delta t## and so on.

I don't disagree with you. I expressed my views and the total energy conservation approach to solving problems in this article.
 
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  • #28
kuruman said:
That coincides with my view.

Precisely. That is why I think that total energy conservation (Case II, post #23) of an isolated multicomponent system is a better way to approach problems of this sort. Then one can write an equation in which the change in total energy is $$\Delta E_{\text{tot}}=\Delta K + \Delta U_{\!g}+\Delta E_{\text{therm}}=0$$ in which ##\Delta E_{\text{therm}} > 0##. That is always the case. Then one has to find a recipe for the change in thermal energy. For a block sliding on the surface, the recipe says ##\Delta E_{\text{therm}} =\mu_kmgd## where all quantities are positive. For current running through a resistor the recipe says ##\Delta E_{\text{therm}} =I^2R\Delta t## and so on.

I don't disagree with you. I expressed my views and the total energy conservation approach to solving problems in this article.
What about non-conservative forces that do positive work (not converting to heat), where do they live in that representation?
 
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  • #29
erobz said:
What about non-conservative forces that do work positive work, where do they live in that representation?
They are roommates in the same space. If I lift a book from the floor to height ##h## at constant speed, the energy balance is written as ##0=\Delta U_{\!g}+\Delta E_{\text{bio}}.## Suppose one chooses a three-component system consisting of Me + Earth + book. The potential energy change of the Earth and book components is ##\Delta U_{\!g} = mgh## and I write the biological energy of the Me component as ##\Delta E_{\text{bio}}.## Then ##0=\Delta E_{\text{bio}}+mgh \implies \Delta E_{\text{bio}}=-mgh. ## Since ##mgh## is positive, this says that the biological energy of Me decreases. System component Me might consider eating something to replenish it.

One can also choose a two-component system consisting of Earth plus book. In that case, Me is external to the system and does work on the book component of the system. The change in internal energy of the system is the change in potential energy of the Earth and book. According to the first law of thermodynamics, the change in internal energy is equal to the external work done by Me (##Q=0.##) Then, ##\Delta U_{\!g}=W_{\text{Me}}\implies W_{\text{Me}}=mgh.## Considering what was said above, Me, an entity that exerts a non-conservative force, does positive work on the Earth-book system and changes the Earth-book system's internal energy at the expense of Me's biological energy.

Finally, one can choose only the book as a one-component system. Then potential energy is meaningless because it takes at least two entities to have a configuration that entails potential energy. The change in kinetic energy of the system is ##\Delta K=0.## The total work done on the system is the sum of the work done by gravity, ##-mgh##, and the work done by Me, ##W_{\text{Me}}##. According to the work-energy theorem ##0=-mgh+W_{\text{Me}}\implies W_{\text{Me}}=mgh.##
 
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  • #30
PeroK said:
It's funny how many question setters think friction acts on a rolling wheel! It should be rolling resistance. The main thing, however, that would slow a bicycle going downhill would be air resistance.

This is not quite right, as you have not treated the slope correctly. If a slope is ##500 \ m## long, then it's ##500 \ m## long. It's not ##500 \cos \theta \ m## long.

There are several ways to solve this problem. Using GPE is one, but not the only one.
You said there were other ways to solve this problem. What are those?
 
  • #31
paulimerci said:
You said there were other ways to solve this problem. What are those?
Uniform acceleration; energy is net force times distance; GPE + friction.
 
  • #32
PeroK said:
It's funny how many question setters think friction acts on a rolling wheel! It should be rolling resistance. The main thing, however, that would slow a bicycle going downhill would be air resistance.

This is not quite right, as you have not treated the slope correctly. If a slope is ##500 \ m## long, then it's ##500 \ m## long. It's not ##500 \cos \theta \ m## long.

There are several ways to solve this problem. Using GPE is one, but not the only on

kuruman said:
They are roommates in the same space. If I lift a book from the floor to height ##h## at constant speed, the energy balance is written as ##0=\Delta U_{\!g}+\Delta E_{\text{bio}}.## Suppose one chooses a three-component system consisting of Me + Earth + book. The potential energy change of the Earth and book components is ##\Delta U_{\!g} = mgh## and I write the biological energy of the Me component as ##\Delta E_{\text{bio}}.## Then ##0=\Delta E_{\text{bio}}+mgh \implies \Delta E_{\text{bio}}=-mgh. ## Since ##mgh## is positive, this says that the biological energy of Me decreases. System component Me might consider eating something to replenish it.

One can also choose a two-component system consisting of Earth plus book. In that case, Me is external to the system and does work on the book component of the system. The change in internal energy of the system is the change in potential energy of the Earth and book. According to the first law of thermodynamics, the change in internal energy is equal to the external work done by Me (##Q=0.##) Then, ##\Delta U_{\!g}=W_{\text{Me}}\implies W_{\text{Me}}=mgh.## Considering what was said above, Me, an entity that exerts a non-conservative force, does positive work on the Earth-book system and changes the Earth-book system's internal energy at the expense of Me's biological energy.

Finally, one can choose only the book as a one-component system. Then potential energy is meaningless because it takes at least two entities to have a configuration that entails potential energy. The change in kinetic energy of the system is ##\Delta K=0.## The total work done on the system is the sum of the work done by gravity, ##-mgh##, and the work done by Me, ##W_{\text{Me}}##. According to the work-energy theorem ##0=-mgh+W_{\text{Me}}\implies W_{\text{Me}}=mgh.#
Thank you, It explains a lot.
I would like to do the problem again, which I posted in #1.
I will take ##\theta## as the angle of incline and ## \phi## as the angle between the force and displacement vectors. I'm using the work energy theorem to solve for the velocity of the biker at the bottom of the hill.
$$ F_\parallel \cdot d cos \phi + F_F \cdot d cos\phi = \frac {1}{2}mv_f^2$$
##\phi## for the first term is zero because the force parallel is along the direction of motion, and ##\phi## for the second term is 180 because the force of friction is opposite to the direction of motion.
$$ mgd sin \theta - F_F \cdot d = \frac{1}{2}mv_f^2$$
$$v_f = 13.7 \frac{m}{s}$$
I did the same thing, but this time I defined what angles I was using and wanted to check whether I did it right this time.
 

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  • #33
PeroK said:
Uniform acceleration; energy is net force times distance; GPE + friction.
Thanks! I'll give it a try on each of them. I think I did the last one, and I made a few mistakes there.
 
  • #34
paulimerci said:
Thanks! I'll give it a try on each of them. I think I did the last one, and I made a few mistakes there.
I tried the first two methods, and I'm getting the correct answer. I don't know how to do the last one.
 
  • #35
paulimerci said:
I tried the first two methods, and I'm getting the correct answer. I don't know how to do the last one.
The kinetic energy of the bicycle at the bottom of the slope equals the gravitational potential energy lost minus the energy lost to friction.

If you do all three methods algebraically, you should see more clearly how they all ultimately involve the same mathematical calculations, just with the key expressions and equations presented differently.
 

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