We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30o. A worker, ignoring friction, calculates that he can do this by giving i

Question

We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30o. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But friction is not negligible; the crateslides only 1.6 m up the ramp, stops, and slides back down.

Find the magnitude of the friction force acting on the crate, assuming that it is constant.

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Philomena 5 years 2021-07-28T03:40:50+00:00 1 Answers 106 views 0

Answers ( )

  1. minhkhang
    0
    2021-07-28T03:42:33+00:00
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    Answer:

    33.72N

    Explanation:

    To find the magnitude of the friction force you take into account all forces over the object:

    -mgsin\theta - F_f+F=ma    (1)

    where Mgsin0 is the gravitational force, Ff is the friction force F is the force of the worker, m is the mass and a is the acceleration of the object while it is going up the ramp.

    it is necessary to find the acceleration of the crate:

    v_f^2=v_o^2-2ad\\\\a\frac{v_f^2-v_o^2}{2d}=\frac{-(5.0m/s)^2}{2(1.6)m}=7.81\frac{m}{s^2}

    with this value and the values for m, g, the angle and d you obtain, by using (1)  you obtain:

    -(12)(9.8)sin30\°-F_f+F=(12)(7.81)\\\\-F_f+F=152.52\\\\(3)

    For the calculation of F:

    If the surface would be frictionless you have:

    -mgsin30\°+F=ma\\\\a=\frac{v_o^2}{2(2.5)m}=5\frac{m}{s^2}

    with these values you can calculate the force F:

    F=ma+mgsin30\°=(12kg)(5m/s^2+9.8m/s^2sin30\°)=118.8N

    BY replacing this values in the expression (3) you get:

    F_f=152.52-118.8N=33.72N

    hence, the magnitude of the friction force is 33.72N

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