A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the object for

Question

A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the object for the next 4.00 s. How far has the object moved during this 7.00 s interval?

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Ngọc Hoa 2 months 2021-07-24T00:24:44+00:00 1 Answers 16 views 0

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    2021-07-24T00:25:54+00:00

    Answer:

    The total distance at 7 s is:

    x_{tot}=27\: m

    Explanation:

    Distance due to the force

    We can use second Newton’s law to find the acceleration.

    F=ma

    a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}

    Now, using the kinematic equation we will find the distance during this interval of time. Let’s recall that the initial velocity is zero.

    x_{1}=0.5at_{1}^{2}

    x_{1}=0.5(1.8)(3)^{2}

    x_{1}=8.1\: m

    In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

    First, we need to find the final velocity of the first interval

    v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s

    So the second distance will be:

    x_{2}=vt_{2}=5.4*4=21.6\: m

    Therefore, the total distance is:

    x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m

    I hope it helps you!

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