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.

Answer 1

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

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

  $v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

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

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

  $v_p=9.35m/s$