In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction betwe

Question

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37â from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart. What are:

a. the speed of the package just before it lands in the cart.
b. the final speed of the cart.

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Adela 4 years 2021-08-13T08:57:30+00:00 1 Answers 291 views 0

Answers ( )

  1. Doris
    0
    2021-08-13T08:58:42+00:00
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    Answer:

    a) v_p=9.35m/s

    Explanation:

    From the question we are told that:

    Open cart of mass   M_o=50.0 kg

    Speed of cart   V=5.00m/s

    Mass of package   M_p=15.0kg

    Speed of package at end of chute V_c=3.00m/s

    Angle of inclination   \angle =37

    Distance of chute from bottom of cart   d_x=4.00m

    a)

    Generally the equation for work energy theory is mathematically given by

      \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

    Therefore

      \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

      v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

      v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

      v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

      v_p=9.35m/s

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