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

a)
`v_p=9.35m/s`

Step-by-step 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

`(1)/(2)mu^2+mgh=(1)/(2)mv_p^2`

Therefore

`(1)/(2)u^2+gh=(1)/(2)v_p^2`

`v_p=\sqrt{2((1)/(2)u^2+gh)}`

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

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

`v_p=9.35m/s`