What is the relationship between power and dissipated energy in cycling?

[ZJacob]

A cyclist is driving along an on an inclined plan with an angle of $\alpha = 0,04$. His mass (including the bike and his stuff) is $75 Kg$.

His speed at an altitude of $1200 m$ (1) is $50 Km/h$ and after some minutes, he pasts by an altitude of $950 m$ (2) with a speed of $62 Km/h$

The image below is an illustration.

1. What is the dissipated energy of friction?

After a certain point, the cyclist velocity stabilises at $v = 45 Km/h$.
2. What is the power of the dissipated energy of friction?

__________

1. The dissipated energy:

$\Delta E_{m} = E_{2} - E_{1} = m ( \frac{ v^{2}_{2} - v^{2}_{1} }{2} + g(z_{2} - z_{1}) ) = - 180 KJ$

Should I say that the dissipated energy is $E = - 180 KJ$ or $E = 180 KJ$ ?

2. The dissipated power:

We know that: $P = \frac{dE}{dt}$

The above equation will become:

$\Delta E_{m} = E_{2} - E_{1} = mg(z_{2} - z_{1}) = mg sin (\alpha)(x_{2} - x_{1})$

Because $v$ is constant.

For small value we take: $sin \alpha = \alpha$, thus: $\Delta E_{m} = E_{2} - E_{1} = mg\alpha(x_{2} - x_{1})$

I don't see how to past to the derivative of $E$ the dissipated energy. I mean, I need to have the expression of $E$ so I can have its derivative.

* Am I allowed to past from $\Delta E$ to $dE$?

* If yes. Dividing by $dt$ is possible, too? so I can have the derivative?

Thank you for your help.

 

[Drakkith]

Hi ZJacob!

I've moved your post to the homework forums. In the future, please make sure you post your homework questions in the appropriate homework forum and use the template provided when creating a post.

 

[ZJacob]

Anyone?

 

[Drakkith]

* Am I allowed to past from $\Delta E$ to $dE$ ?

Since the velocity has stabilized at 45 km/h, is the energy dissipated per unit of time changing? If you were to graph the energy dissipated over time, what would that look like? How is $\frac{dE}{dt}$ related to that graph?

 

[ZJacob]

Since the velocity has stabilized at 45 km/h, is the energy dissipated per unit of time changing? If you were to graph the energy dissipated over time, what would that look like? How is $\frac{dE}{dt}$ related to that graph?

* The energy dissipated per unit of time is not changing after the velocity has stabilized since: $\Delta E_{m} = mg\alpha v$.

* If I were to graph the energy dissipated, the graph would look like this:

* $\frac{dE}{dt}$ is the slope.

Q: Could you please answer my first question about the sign of the dissipated energy? Thank you.

 

[Drakkith]

* $\frac{dE}{dt}$ is the slope.

So what's the slope of a straight line and how does that relate to $\frac{ΔE}{Δt}$?

Q: Could you please answer my first question about the sign of the dissipated energy? Thank you.

I would if I could.

 

[ZJacob]

So what's the slope of a straight line and how does that relate to $\frac{ΔE}{Δt}$?

I would if I could.

The slope of a straight line is: $\frac{\Delta f}{\Delta x}$. Also $\frac{ΔE}{Δt}$ represent the rate of change of the energy. Which can also be written as: $\frac{dE}{dt}$. So, I'm allowed to past from $\Delta E_{m}$ to $dE_{m}$. Correct?

Why? It would be wrong if I wrote: The energy dissipated is: $E = - 180 KJ$ ? Thank you.

 

[Drakkith]

So, I'm allowed to past from $\Delta E_{m}$ to $dE_{m}$ . Correct?

As far as I know, yes.

Why? It would be wrong if I wrote: The energy dissipated is: $E=−180KJ$?

I'm sorry but I don't know for certain.

 

[ZJacob]

As far as I know, yes.
I'm sorry but I don't know for certain.

Thank you.