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12 April, 23:39

A car drives 5.0 km north, then 7.3 km east, then 5.3 km northeast, all at a constant velocity. If the car had to perform 2.6 * 106 J of work during this trip, what was the magnitude of the average frictional force on the car?

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Answers (2)
  1. 13 April, 00:33
    0
    147.73 N

    Explanation:

    Since Work, W = Force, F * distance, d

    W = Fd

    F = W/d W = 2.6 * 10⁶ J and d = 5.0 km + 7.3 km + 5.3 km = 17.6 km = 17600 m

    F = W/d = 2.6 * 10⁶ J/17600 m = 147.73 N. Which is the force causing the car to move.

    Now if f is the average frictional force on the car,

    F - f = ma

    Since the car moves at constant velocity, a = 0

    F - f = 0

    F = f = 147.73 N
  2. 13 April, 02:33
    0
    147.73N

    Explanation:

    Work is said to be done if an applied force causes a body to move through a distance.

    Work done by the car = Force * distance.

    Given:

    Work done = 2.6 * 10^6 Joules

    Total distance covered by the car = 5.0km + 7.3km + 5.3km

    = 17.6km

    = 17600m

    From the formula:

    Force = Work done/Total distance

    Force = 2.6*10^6/17600

    Force experienced by the car = 147.73N

    Magnitude of the average frictional force on the car is 147.73N
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