A force is applied to a block to move it up a 30° incline. The incline is frictionless, the block is initially at rest, F= 65.0 N and M=5.00

Question

A force is applied to a block to move it up a 30° incline. The incline is frictionless, the block is initially at rest, F= 65.0 N and M=5.00 kg. If the block starts from rest, how far will it move in 0.550 s? Also, indicate whether the block moves up or down the incline.

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Phúc Điền 5 months 2021-09-05T12:10:54+00:00 1 Answers 11 views 0

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    2021-09-05T12:12:47+00:00

    Answer:

    0.96 m, upward

    Explanation:

    \theta=30^{\circ}

    F=65 N

    M=5 kg

    Initial velocity, u=0

    t=0.550 s

    Fcos\theta-Mgsin\theta=Ma

    Where g=9.8 m/s^2

    Substitute the values

    65cos30-5\times 9.8sin30=5a

    31.8=5a

    a=\frac{31.8}{5}=6.36 m/s^2

    S=ut+\frac{1}{2}at^2

    S=0+\frac{1}{2}(6.36)(0.55)^2=0.96 m

    Hence,the block moves upward because displacement is positive.

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