Question

A 5.0 kg mass is added and the mass-spring system is turned horizontal. If the spring is displaced to - 2.0 cm, what is the potential energy of the mass-spring system when released just before it starts to move?

(a) What is the maximum speed of the 5.00-kg mass?
cm/s
(b) What is the maximum acceleration of the 5.00-kg mass?
cm/s2
(c) What are the position, velocity, and acceleration of the 5.00-kg mass as functions of time? (Use the following as necessary: t. Do not enter units in your expression. Assume t is in seconds.)

Answer 1

The following answers assumed the initial hanging load on the spring was 50N and that the spring was stretched by this load a distance of 5.0cm this gives a spring constant of 1000N/m

Potential Energy U = 0.2J

(a) Vmax = 0.28m/s

(b) amax = 4m/s²

(c) Vx = -14.14Sin(14.14t + phi)

ax = -200Cos(14.14t + phi)

Step-by-step explanation:

The detailed solution is contained in the attachment below. The solutions have been pressed in a step by step approach to all the answers.

The potential energy in a spring that has been compromised or stretched a distance is equal to U

= ½kx²

In this case the spring was compressed a distance x = 2.0cm

Vmax = The maximum velocity when the net sum of displacement made by the spring is equal to zero. (When the spring passes through the equilibrium point)

Vmax (K/m)½ * A

amax = -k/m * A

Tgank you foe reading and I hope this was helpful to you.