g = 9.8 m/s^2 Earth surface gravity 1 mi = 1069 m 1 km = 10^3 m 1 hr = 3600 s 1
ID: 2075612 • Letter: G
Question
g = 9.8 m/s^2 Earth surface gravity 1 mi = 1069 m 1 km = 10^3 m 1 hr = 3600 s 1 g = 10^-3 kg A friend of yours says that when you lift an object off the ground, the object gains gravitational potential energy. Why is your friend incorrect? How would you rephrase your friend's claim so that it is more accurate? Which categories of energy are effected by irreversible processes but unaffected by reversible processes? Briefly explain why that is the case. A 2000 kg car has a fuel economy of 30 mi/gal meaning that in an typical 30 mile trip the car burns through one gallon of gasoline. Our gallon of gasoline stores about 125 million Joules of source energy. You want to figure out what fraction of the car's fuel is used to actually keep the car moving. In other words, you want to figure out the car's energy efficiency. You get the car up to a cruising speed of 60 mi/hr. You release the accelerator pedal completely and find the car's speed drops by about 0.75 mi/hr in one second. How much kinetic energy does the car lose in one second Assume the value calculated in part (d1) equals the amount of useful mechanical energy generated by your engine every second in order to maintain a steady 60 mi/hr speed. How much mechanical energy does your engine generate for a 30 mile trip? Based on your calculations and the given information about the amount of source energy stored in gasoline, estimate the energy efficiency of your car. How much of the source energy is lost to thermal energy during a 30 mile trip?Explanation / Answer
A) Gravitational potentional energy U=mgh
where as,
m--->mass of the object
g--->accelaration due to gravity
h---->hight of the object
now think about the sign of g&h since both are vector quntity .both are oposit in diraction so the product will always -ve so by increasing 'h' the magnitude of 'U' will increase but due to its negative sign potential is actully decreasing.
B) The totel Mechanical energy will not effected by a reversible process.but it change in reversible
C1) v0=60mi/hr=26.82m/s
initial K.E=1/2*mv2
=0.5*2000*26.822
=719312.4 J
v=.75mi/hr=0.34m/s
final K.E=0.5*2000*.342=115.6 j
energy lose=719312.4-115.6=719196.8 J
C2) let assume there is no potential energy change and frictional losess
the mechanical energy created is only its kinetic energy so the totel kinetic energy created
K.E=0.5*2000*26.822=719312.4 J
C3) efficacy=work done/energy used
=719312.4/125000000=0.58% (look at the numbers..!!!)
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