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18 March, 18:24

4. The fire alarm goes off, and a 97 kg fireman slides 3.0 m down a pole to the ground floor. Suppose the fireman starts from rest, slides with a constant acceleration, and reaches the ground floor in 1.2 s. (A) What was the upward force F exerted by the pole on the fireman

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  1. 18 March, 21:22
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    The upward force F was 546.11 N

    Explanation:

    In order to find the value of the force F, you have to apply Newton's Second Law:

    ∑F=ma

    where F represents all the forces, m is the mass and a is the acceleration.

    In this case, the applied forces are the force F exerted by the pole and the weight force, due to the acceleration of gravity.

    Replacing in the formula the components of the forces in the y-axis:

    Using a positive direction downward:

    -F + W = ma

    -F + mg=ma

    F = mg - ma (I)

    Notice that F is upward and the weight is downward.

    You need to calculate the acceleration with the given information and then you can calculate the value of F.

    Applying kinematic motion formula:

    Yf = Yo + Vot + 0.5at²

    where yo is the initial position, vo is the initial velocity, a is the acceleration, t is the time and Yf is the final position.

    In this case:

    t=1.2s

    vo=0 (because the fireman starts from rest)

    Yo = 0 m and Yf = 3.0 m

    Therefore:

    3.0 = 0 + 0 (1.2) + 0.5a (1.2) ²

    3.0 = 0.72a

    Dividing by 0.72

    a = 4.17 m/s²

    Replacing it in (I):

    F = 97 (9.8) - 97 (4.17)

    F = 546.11 N
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