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A 1500-kg car travels on a level road at 42 m/s (State 1). It then continues down a hill to a lower part of the road (State 2), dropping in elevation by 25 meters in the process.If the car coasts without friction, what is its speed in State 2? Assume that g = 9.81 m/s2.Now imagine it coasts, starting at 42 m/s in State 1, but there is friction from aerodynamic drag and rolling resistance. The car’s speed in State 2 is now 43 m/s. If the car’s temperature stays constant, what is the heat transfer to the surrounding air?

Respuesta :

Answer:

speed in state 2 is 47.48 m/s

heat transfer is 304.125 kJ

Explanation:

given data

mass m1 = 1500 kg

velocity v1 = 42 m/s

elevated h1 = 25 m

velocity v2 = 42 m/s

velocity v3 = 43 m/s

to find out

speed in State 2 and heat transfer to surrounding

solution

we know that total energy always remain constant so we can say

KE + PE = KE1 + PE1

so

1/2×  mv1²  + mgh = 1/2×  mv²  + mgh1

1/2×  1500(42)²  + 0  = 1/2×  1500v²  + 1500 (9.81) 25

solve and we get v

v = 47.48 m/s

speed in state 2 is 47.48 m/s

and

we know energy loss by the drag in form of heat

so

heat transfer is

heat transfer = 1/2 × m × ( v - v3 )²

heat transfer = 1/2 × 1500 × ( 47.48 - 43 )²

heat transfer is 304.125 kJ

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