Calculate Work by Non-Conservative Forces for 1.10x10^3kg Car

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AI Thread Summary
The discussion focuses on calculating the work done by non-conservative forces on a 1.10 x 10^3 kg car moving up a mountain road. The car does 6.80 x 10^6 J of work while reaching a speed of 22.0 m/s at an altitude of 190 m. The user initially calculates the work done by the car as 2,335,960 J but realizes a mistake in their calculations. The correct approach reveals that the work done by non-conservative forces, accounting for friction and air resistance, results in a loss of energy totaling -4,464,040 J. The discussion highlights the importance of accurately applying the work-energy principle in physics problems.
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Homework Statement



The (non-conservative) force propelling a 1.10 x 10(3)-kg car up a mountain road does 6.80 x 10(6) J of work on the car. The car starts from rest at sea level and has a speed of 22.0 m/s at an altitude of 1.90 x 10(2) m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non-conservative forces.
Calculate the work done by non-conservative forces using Wnc = dKE +DPE where
Wnc= W friction+W air+W engine

Homework Equations


I used W1= .5MV2 + MGH
and
required W =W1-W


The Attempt at a Solution



.5*1100kg*(22squared m/s) + 1100kg*9.8*192m = 2335960J

2335960-6800000 = -1655960J
What did I do wrong?
 
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Work done on the car - work done by the car = loss of energy due to friction, air resistance.
 
the 2335960-6800000 gives -4464040 which is the correct answer. Too tired to see how unreasonable the first result was!
 
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