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28 March, 19:03

If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a time period of 5 sec,

Determine:

1) What is the net force acting on it?

2) Acceleration acting upon the box if it starts from rest and attains a certain Kinetic energy after being pushed 25m?

3) Also find the mass of the box.

4) What will be the kinetic energy and final velocity?

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Answers (1)
  1. 28 March, 19:20
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    The force acting in the front direction is the 130N.

    The frictional force is acting backwards 30N.

    1) The net force is 130N - 30N = 100N

    2) s = ut + (1/2) at^2 u = 0, Start from rest, s = 25m t = 5.

    25 = 0*5 + (1/2) * a * 5^2.

    25 = 0 + 25/2 * a.

    25 = (25/2) a. Divide 25 from both sides.

    1 = (1/2) * a. Cross multiply.

    2 = a.

    a = 2 m/s^2.

    3) Mass of the box

    Net Force, F = ma

    100 = m*2. Divide both sides by 2.

    100/2 = m

    50 = m.

    m = 50 kg.

    4) Final velocity, v = u + at.

    v = 0 + 2*5 = 10 m/s.

    Kinetic Energy, K = (1/2) * mv^2.

    = 1/2 * 50 * 10 * 10.

    = 2500 J.
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