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?

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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