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31 December, 11:04

5. A 1052 kg truck, starting from rest, reaches a speed of 20.0 m/s in 6.20 s.

a. What is the average acceleration of the truck?

b. What is the magnitude of the average net force on the truck during the 6.20 s?

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Answers (1)
  1. 31 December, 13:30
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    a. 3,392.7 N

    b. 3,392.7 N

    Explanation:

    We are given the following information;

    Mass of the truck as 1052 kg initial speed as 0 m/s Final speed as 20.0 m/s Time taken as 6.20 s

    #a. We are required to calculate the acceleration;

    We need to know the formula of getting acceleration;

    a = (v-u) / t

    Where v is the final velocity, u is the initial velocity

    Therefore;

    a = (20 m/s - 0 m/s) / 6.20s

    = 3.225 m/s²

    Thus, the average acceleration of the truck is 3.225 m/s²

    #b. We are required to calculate the net force on the truck

    We need to know that;

    According to the second Newton's law of motion, F=ma

    Where F is the net force, m is the mass and a is the acceleration.

    Therefore;

    Net force, F = mass * Acceleration

    = 1052 kg * 3.225 m/s²

    = 3,392.7 N

    Thus, the net force on the truck is 3,392.7 N
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