- #1
ZJacob
- 5
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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.
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.