A car of mass 1000 kg is at the top of a degree hill as shown. What is its gravi

Question

A car of mass 1000 kg is at the top of a degree hill as shown. What is its gravitational potential energy relative to the bottom of the hill? U = 1000kg times 9.8m/s2 times 50m tan(10) = 8.6 t times 104 J U = 1000kg times 50m = 5.0 times 105 J U = 1000 times 9.8 times 50 = 4.9 times 105 J U = 1000kg times 9.8m/s2 times 50m times 10 = 4.9 times 106 J U = 1000kg times 9.8m/s2 times 50m times sin(10) = 8.5 times 104 J If the car rolls down the hill ( with the engine off) with negligible friction and air resistance, what will its kinetic energy be when it reaches the be when it reaches the bottom of the hill? Suppose instead that the amount of work done on the car by the car by frictional and air resistance forces as the car rolls down the hill is 50,000 J. What then is the kinetic energy of the car when it reaches the bottom of the hill?

Explanation / Answer

1. potential energy at top = mgh = 1000*9.8*50*tan(10) = 8.6*10^4J potential energy at bottom = 0 so relative potential energy = 8.6*10^4J answer is 1 2. since there is no external force energy will be conserved and all potentail energy at top will be converted into kinetic energy at bottom so kinetic energy = 8.6*10^4 J 3. when work done is 50,000 J so kinetic energy = 86000 - 50000 = 360000 J = 3.6*10^4 J

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