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

[PeroK]

You said there were other ways to solve this problem. What are those?

Uniform acceleration; energy is net force times distance; GPE + friction.

 

[paulimerci]

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

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.

 

[paulimerci]

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.

 

[paulimerci]

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.

 

[PeroK]

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.

 

[paulimerci]

I get it now. Yes, I see that a single problem can be approached in many different ways and ultimately give the same result. I can now understand more clearly. Was my approach to post #32 right?

 

[PeroK]

I get it now. Yes, I see that a single problem can be approached in many different ways and ultimately give the same result. I can now understand more clearly. Was my approach to post #32 right?

It looks right, but a bit overcomplicated IMO. One good thing about a free-body diagram is that you can label the forces with an arrow showing the direction of each force. This allows you to jump in with the algebra already simplified somewhat. If $f$ is the friction force acting up the slope and $l$ is the length of the slope, then you can simply write$$F_{net} = mg \sin \theta - f$$or$$W_{net} = (mg \sin \theta - f)l$$You don't have to invoke angles of $\phi =0$ and $\phi = \pi$ etc.

There's something to be said for keeping things simply initially until you develop some intuition for these problems.

 

[paulimerci]

It looks right, but a bit overcomplicated IMO. One good thing about a free-body diagram is that you can label the forces with an arrow showing the direction of each force. This allows you to jump in with the algebra already simplified somewhat. If $f$ is the friction force acting up the slope and $l$ is the length of the slope, then you can simply write$$F_{net} = mg \sin \theta - f$$or$$W_{net} = (mg \sin \theta - f)l$$You don't have to invoke angles of $\phi =0$ and $\phi = \pi$ etc.

There's something to be said for keeping things simply initially until you develop some intuition for these problems.

Thank you; your approach looks simple and easy to understand.

 

[paulimerci]

I appreciate @kuruman @PeroK @erobz @jbriggs444 taking the time to thoroughly explain this idea. I'm incredibly grateful for that.