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8 May, 14:05

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial velocity of 2.77 m/s. The car then runs up the frictionless slope, gaining 0.184 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance?

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  1. 8 May, 17:31
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    The final velocity of the car is 2.02 m/s

    Explanation:

    Hi there!

    The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:

    KE = 1/2 · m · v²

    Where:

    KE = kinetic energy

    m = mass

    v = velocity

    Then, the initial kinetic energy will be:

    KE = 1/2 · 0.100 kg · (2.77 m/s) ²

    KE = 0.384 J

    When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let's calculate the potential energy of the car as it reaches the top:

    PE = m · g · h

    Where:

    PE = potential energy.

    m = mass

    g = acceleration due to gravity.

    h = height.

    PE = 0.100 kg · 9.8 m/s² · 0.184 m

    PE = 0.180 J

    Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J

    Using the equation of kinetice energy, we can obtain the velocity of the car:

    KE = 1/2 · m · v²

    0.204 J = 1/2 · 0.100 kg · v²

    2 · 0.204 J / 0.100 kg = v²

    v = 2.02 m/s

    The final velocity of the car is 2.02 m/s
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