A package delivery service (we can\'t say any names but their initials are F and

Question

A package delivery service (we can't say any names but their initials are F and E), has asked your help in designing a system to bring packages that have been sliding along delivery chutes before they get loaded onto the trucks. The packages, with a mass of 5 kg, are stationary at the top and then they slide down a chute that is 5m--high. The delivery chute has rollers along it which make it virtually frictionless. The delivery chute makes and angle of 25.9 degrees. A stopping spring is located at the very end of the entire chute but the -3.8m -wide horizontal slowing chute between the bottom of the delivery chute and right in front of the spring does not have rollers and is not frictionless (uk=.21). The stopping spring has a spring constant of 490 N/m. For simplicity, assume the chute under the spring is frictionless, otherwise the math gets ugly.

What is the speed of the package at the bottom of the chute? in m/s

What is the speed of the box just before hitting the stopping spring? in m/s

How far is the stopping spring compressed as it brings the package to a stop? in m

Explanation / Answer

KE at the bottom of chute = mgh = 5*9.8*5*sin25.9 = 107.01 J

v at bottom:

.5mv^2 = 107.01

v = 6.54 m/sec

final v after frictional chute:

v'^2 = v^2 - 2*.21*g*3.8

v' = 5.21 m/sec

now, kinetic energy when block reaches spring:

KE = .5mv'^2 = 67.91 J

this energy will be used to compress spring:

.5k*x^2 = 67.91

x = .526 m

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