Thermal Energy dissapated from brakes.

AI Thread Summary
To calculate the thermal energy dissipated from the brakes of a 1600 kg car descending a 20-degree hill, the conservation of energy equation is applied. The key is to consider the change in kinetic and potential energy, ensuring to convert speed from km/h to m/s for accurate calculations. It's important to use different symbols for velocity and height on each side of the equation to avoid confusion. The discussion highlights the significance of unit conversion and proper variable representation in solving the problem. Properly applying these principles leads to the correct determination of energy loss during braking.
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This is another one that I have given several attempts.

Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 20 degree hill. The car begins braking when its speed is 95 km/h and slows to a speed of 30 km/h in a distance of 0.27 km measured along the road.

I tried using the conservation of energy equation.

1/2mv^2+mgh=1/2mv^2+mgh+Eloss

This doesn't seem to give me what I'm looking for.

Any help is greatly appreciated
 
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You have the correct method.
Think of it terms of speed and kinetic energy change.
change in ke + change in pe = energy loss.
 
Just remember to use m/s instead of km/h, and make sure you get the right value for the change in height of the car.

In your equation, the v and h on the rhs should be different from those on the lhs. Don't use the same symbol for different quantities!
 
Thank you soooo, much. The problem was just that I had the velocity in the wrong units.
I would have spent forever on that problem!
 
Excellent!
 
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