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15 January, 02:43

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?

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  1. 15 January, 03:11
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    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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