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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.

Sagot :

okpalawalter8

Answer:

a) [tex]v_p=9.35m/s[/tex]

Explanation:

From the question we are told that:

Open cart of mass   [tex]M_o=50.0 kg[/tex]

Speed of cart   [tex]V=5.00m/s[/tex]

Mass of package   [tex]M_p=15.0kg[/tex]

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

Angle of inclination   [tex]\angle =37[/tex]

Distance of chute from bottom of cart   [tex]d_x=4.00m[/tex]

a)

Generally the equation for work energy theory is mathematically given by

  [tex]\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2[/tex]

Therefore

  [tex]\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2[/tex]

  [tex]v_p=\sqrt{2(\frac{1}{2}u^2+gh)}[/tex]

  [tex]v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}[/tex]

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

  [tex]v_p=9.35m/s[/tex]

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