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    Newton's Laws of Motion : Inclined Plane

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    A 14 kg. crate starts at rest at the top of a 50 degrees incline. The coefficients of friction are us=0.38 and uk=0.29. The crate is connected to a hanging 8.2 kg box by an ideal rope and pulley.
    (a) As the crate slides down the incline, what is the tension in the rope?
    (b) How long does it take the crate to slide 1.94 m down the incline?
    (c) To push the crate back up the incline at constant speed, with what force P should Pauline push on the crate (parallel to the incline)?

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    Solution Preview

    (Please refer attachment for fig.)

    a) Fig. shows all the forces acting on the system. Weight 14g of the crate has been resolved into components 14gsin50 and 14gcos50. N is the normal reaction of the inclined plane on the crate.

    N = 14gcos50 ..........(1)

    f is the frictional force on the crate due to the inclined plane.

    f = uk x N = uk x 14gcos50 = 0.29 x 14 x 9.81 x 0.64 = ...

    Solution Summary

    Step-by-step solution regarding inclined plane is provided.