Net Power to Accelerate Car up Inclined Road

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To find the net power required for a car to accelerate up an inclined road to a height of 44 meters, the work-energy principle is applied, focusing on the change in kinetic and gravitational potential energy. The net work done on the car is equal to the change in kinetic energy, calculated as 1/2 * m * (V_f^2 - V_i^2), where V_i is 0. Additionally, the gravitational work must be considered, as the car is moving uphill, which adds to the total energy required. The correct setup of the energy equation leads to a total energy requirement of approximately 4.9 x 10^5 joules. Understanding how to incorporate velocity and time into the equations is crucial for accurate calculations.
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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:
\Delta E = W_{net}
 
I still don't get it after that...
 
Can you set up the energy equation? You should get about 4.9 \times 10^5 joules.
 
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