# Homework Help: Power of dissipated energy.

1. Dec 15, 2016

### 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?

2. Dec 15, 2016

### Drakkith

Staff Emeritus
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.

3. Dec 16, 2016

### ZJacob

Anyone?

4. Dec 16, 2016

### Drakkith

Staff Emeritus
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?

5. Dec 16, 2016

### ZJacob

* 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.

6. Dec 16, 2016

### Drakkith

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

I would if I could.

7. Dec 16, 2016

### ZJacob

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.

8. Dec 16, 2016

### Drakkith

Staff Emeritus
As far as I know, yes.

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

9. Dec 16, 2016

Thank you.