Work energy principle and power

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Discussion Overview

The discussion revolves around calculating the average power generated by a car's engine as it accelerates up an inclined hill, using the work-energy principle. Participants explore various methods to arrive at the solution, including calculations of acceleration, forces, and work done, while addressing discrepancies between their results and a textbook answer.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving a car's mass, incline, initial and final speeds, and requests help in finding the average power, noting a discrepancy with the textbook answer of 36000 W.
  • Another participant questions the original poster's understanding and suggests starting with a diagram and the work-energy theorem.
  • A participant calculates an acceleration of 0.033 m/s² and a force of 2831.7 N, but arrives at an average power of 31180 W, differing from the textbook answer.
  • Another participant provides calculations for average velocity and total work done, proposing two methods to calculate average power, but also does not arrive at the textbook answer.
  • Participants share formulas for kinetic energy and gravitational potential energy, indicating different approaches to the problem without agreeing on a final answer.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct average power value, with multiple competing calculations and methods presented. Discrepancies between individual results and the textbook answer remain unresolved.

Contextual Notes

Participants express uncertainty regarding their calculations and the assumptions made in their approaches, including the treatment of forces and energy changes. Some calculations depend on specific interpretations of the work-energy theorem.

Shah 72
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A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.
The and in the textbook is 36000 W, iam not able to do. Pls help
 
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Shah 72 said:
A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.
The and in the textbook is 36000 W, iam not able to do. Pls help
Again what have you been able to do? You've posted a large number of these questions and have learned nothing about how to start? How about a diagram and a statement of the work energy theorem? See what you can do, then if you still are getting an incorrect answer post what you have done and we can look it over.

-Dan
 
topsquark said:
Again what have you been able to do? You've posted a large number of these questions and have learned nothing about how to start? How about a diagram and a statement of the work energy theorem? See what you can do, then if you still are getting an incorrect answer post what you have done and we can look it over.

-Dan
Iam getting the ans 31180W, but textbook ans is 3600W
I first calculated acceleration using v= u+at and got a= 0.033m/s^2
Then calculated F- 16000sin10=1600×0.033 and got the ans F=2831.7N
Then calculated s using v^2= u^2+2as and got s= 660.67m
Avg power= work done/ time
Iam not getting the right ans. Pls help
 
A car of mass 1200 kg accelerates up a hill inclined at 10 degree to the horizontal. The car has initial speed of 10 m/ s and final speed of 12 m/s after 60 s. Air resistance and friction may be ignored. Find the average power generated by the engine.

Shah 72 said:
Iam getting the ans 31180W, but textbook ans is 3600W

Well, I'm not getting either of those values. First off, some basic values ...

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{1}{30} \, m/s^2$

$v_{avg} = \dfrac{v_0+v_f}{2} = 11 \, m/s$

$\Delta x = v_{avg} \cdot \Delta t = 660 \, m$method 1

$P_{avg} = \dfrac{\text{total work done by the engine}}{\text{elapsed time}}$

$P = \dfrac{\Delta KE + \Delta GPE}{\Delta t}$

$\color{red} P = \dfrac{\frac{1}{2}m(v_f^2-v_0^2) + mg \Delta x \cdot \sin{\theta}}{\Delta t}$

method 2

$P_{avg} = (\text{force exerted by the engine})(\text{avg velocity}) = F_e \cdot v_{avg}$

$F_e - mg\sin{\theta} = ma \implies F_e = m(g\sin{\theta} + a)$

$\color{red} P = m(g\sin{\theta}+a) \cdot v_{avg}$
 
skeeter said:
Well, I'm not getting either of those values. First off, some basic values ...

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{1}{30} \, m/s^2$

$v_{avg} = \dfrac{v_0+v_f}{2} = 11 \, m/s$

$\Delta x = v_{avg} \cdot \Delta t = 660 \, m$method 1

$P_{avg} = \dfrac{\text{total work done by the engine}}{\text{elapsed time}}$

$P = \dfrac{\Delta KE + \Delta GPE}{\Delta t}$

$\color{red} P = \dfrac{\frac{1}{2}m(v_f^2-v_0^2) + mg \Delta x \cdot \sin{\theta}}{\Delta t}$

method 2

$P_{avg} = (\text{force exerted by the engine})(\text{avg velocity}) = F_e \cdot v_{avg}$

$F_e - mg\sin{\theta} = ma \implies F_e = m(g\sin{\theta} + a)$

$\color{red} P = m(g\sin{\theta}+a) \cdot v_{avg}$
Thank you!
 

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