Work energy principle and power

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SUMMARY

The discussion focuses on calculating the average power generated by a car engine while accelerating up a 10-degree incline. The car has a mass of 1200 kg, an initial speed of 10 m/s, and a final speed of 12 m/s over a duration of 60 seconds. The correct average power, as confirmed by calculations, is 36,000 W, derived from the work-energy principle, which includes both kinetic and gravitational potential energy changes. Participants provided various methods for calculating average power, emphasizing the importance of using correct formulas and values.

PREREQUISITES
  • Understanding of the work-energy theorem
  • Basic knowledge of kinematics, including acceleration and average velocity
  • Familiarity with gravitational potential energy calculations
  • Ability to perform vector calculations involving forces on inclined planes
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn how to calculate gravitational potential energy on inclines
  • Explore kinematic equations for uniformly accelerated motion
  • Practice problems involving forces on inclined planes and average power calculations
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the principles of work, energy, and power in real-world applications.

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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