Net Power to Accelerate Car up Inclined Road

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Homework Help Overview

The discussion revolves around calculating the net power required for a car to accelerate up an inclined road to a specified height and speed within a given time frame. The subject area includes concepts from mechanics, specifically work-energy principles and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the work-energy principle but expresses confusion regarding the inclusion of terms related to the incline and the energy changes involved. Participants suggest setting up the energy equation and highlight missing components in the original poster's approach.

Discussion Status

The discussion is ongoing, with participants providing guidance on the formulation of the energy equation. There is an indication that some numerical values have been discussed, but no consensus or resolution has been reached yet.

Contextual Notes

Participants note the neglect of frictional losses in the problem setup, which may influence the calculations. The original poster is also questioning how to incorporate velocity and time into their equations.

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A car (m = 760 kg accelerates uniformly from rest up an inclined road which rises uniformly, to a height, h = 44 m. Find the net power the engine must deliver to reach a speed of 20.7 m/s at the top of the hill in 22.5s (NEGLECT frictional losses: air and rolling, ...)

I know that W_n=Triangle_K (change in K) which is 1/2*m*(V_f^2-V_i^2) where V_i = 0
and I also have that W_g+W_f=1/2 m

I tried solving with that , but I don't get it right

What am I doing wrong?; Also what do I use Velocity and Time for?( I know what for , but where in the equations do I use it if it is to be used).
 
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You're missing some of the terms in your work-energy equation since the car is going uphill.

The equation should be:
[tex]\Delta E = W_{net}[/tex]
 
I still don't get it after that...
 
Can you set up the energy equation? You should get about [tex]4.9 \times 10^5[/tex] joules.
 

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