Question 1066943: Your kindly, old professor drives a Toyota Avalon to work to teach his beloved students. It is a hybrid car that uses an internal combustion engine in conjunction with electric motors and a high-voltage battery for propulsion. The battery/electric motor combination is also used to brake the car by absorbing the kinetic energy of the cars motion and storing that energy in the battery for later use(rather than dissipating it as heat).
The kinetic energy of the cars motion is given by Ek=Wv^2/2g where Ek is the kinetic energy of motion (in joules), W is the weight of the car and its contents (in Kilograms),v is the velocity of the car (in meters per second), and g=9.8 meters per second (the acceleration due to gravity).
The battery voltage is 200 volts. The car is moving at a velocity of 60 mph (26.28 meters per second) when the breaking system is engaged. The car stops in 20 seconds. The car and its contents weigh 1500 kilograms. How much charge is stored in the battery during the stop?
The equations involved are: I=Q/t where I is the battery current (in amperes), Q is the charge stored in the battery during the stop (in coulombs), P=Ek/t where P is the power delivered to the battery during the stop (in watts), and P=IV where V is the battery voltage (in Volts).
Answer by ikleyn(52748)   (Show Source):
You can put this solution on YOUR website! .
Your kindly, old professor drives a Toyota Avalon to work to teach his beloved students. It is a hybrid car that uses an internal
combustion engine in conjunction with electric motors and a high-voltage battery for propulsion.
The battery/electric motor combination is also used to brake the car by absorbing the kinetic energy of the cars motion
and storing that energy in the battery for later use(rather than dissipating it as heat).
The kinetic energy of the cars motion is given by Ek=Wv^2/2g where Ek is the kinetic energy of motion (in joules),
W is the weight of the car and its contents (in Kilograms), v is the velocity of the car (in meters per second),
and g=9.8 meters per second (the acceleration due to gravity).
The battery voltage is 200 volts. The car is moving at a velocity of 60 mph (26.28 meters per second) when the breaking system
is engaged. The car stops in 20 seconds. The car and its contents weigh 1500 kilograms. How much charge is stored
in the battery during the stop?
The equations involved are: I=Q/t where I is the battery current (in amperes), Q is the charge stored in the battery during the stop
(in coulombs), P=Ek/t where P is the power delivered to the battery during the stop (in watts), and P=IV where V is the battery
voltage (in Volts).
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0. You just presented everything or ALMOST everything in your post, so you ONLY need to establish the order (the sequence) of calculations.
This sequence is as follows:
1. Calculate the kinetic energy  =  =  = 517978.8 Joules.
Notice. My opinion is that you make a mistake calling 1500 kg as a weight and dividing by "g" then.
1500 kg is the MASS of the car, while the weight is 1500*9.81 newtons.
Probably (again my personal opinion) it is yours major mistake in the condition, forcing you to post it to this forum
for the clarification.
2. Then calculate the power P =  =  = 25899 watts.
3. Then calculate the battery current I =  =  = 129.5 amperes.
4. Then calculate the charge stored in the battery during the stop
Q = I*t = 129.5*20 = 2590 coulombs.
Answer. The charge stored in the battery during the stop is about 2590 coulombs (approximately).
The only weak place in your understanding is mixing kilograms of mass with kilograms of weight.
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Kilograms of weight were in use till fifties, when the engineers understood EVERYTHING what they were doing.
In sixties it was gone, and kilograms of weight were replaced by newtons.
The unit system SI came in use, which separated these units.
After that kilograms of weight are not in use at all.
(except of those who keep them in their minds since the fifties and still understand what they are doing . . . )
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