Simple Harmonic Motion Question

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A 4.5 KG mass attached to a spring with a constant of 75 KN/m is compressed 0.4 m and then released with a 2.5 KG mass placed in front of it. The masses will separate when the spring force equals the weight of the 2.5 KG mass, calculated to occur at 1.22 m from equilibrium. The 4.5 KG mass will not stop if it moves horizontally due to the lack of friction. When considering energy conservation, the distance between the masses when the 4.5 KG mass stops is determined to be 0.86 m. Accurate unit conversion is crucial for consistent calculations throughout the problem.
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I am going brain dead on a rather simple question. Appreciate some help.


A mass of 4.5 KG attatched to a horizontal spring of constan 75 KN/m is compressed 0.4 m from Equilibirium and then a loose mass of 2.5 KG is placed in front of it . IF the mass is now released:

1. at what point do the masses separate.
2. How far apart are they when the 4.5 KG mass stops?


Assume no friction.
 
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part 1:
the mass moves at the same speed with the spring while they are in contact, so, they will separate at the point when the spring started to slow down (deccelarate) [assuming the spring is ideal, massless, and ignore the effect of gravity]
part 2:
the mass will not stop if it shoots horizontally... coz you assume no friction...
if it shoots vertically, you can use conservation of energy... the energy stored in spring is 1/2 kx^2
 


1. The masses will separate when the spring force acting on the 4.5 KG mass is equal to the weight of the 2.5 KG mass. This can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium. Plugging in the given values, we get F = (75 KN/m)(0.4 m) = 30 KN. Since the weight of the 2.5 KG mass is 24.5 N, the masses will separate when the displacement of the 4.5 KG mass is 30 KN/24.5 N = 1.22 m from equilibrium.

2. To find the distance between the masses when the 4.5 KG mass stops, we need to use the conservation of energy principle. At the point where the 4.5 KG mass stops, all the potential energy stored in the spring will be converted into kinetic energy of the masses. We can set up the equation as follows:

1/2kx^2 = 1/2mv^2

Where k is the spring constant, x is the displacement of the 4.5 KG mass, m is the total mass (4.5 KG + 2.5 KG = 7 KG), and v is the velocity of the masses. Solving for x, we get x = √(mv^2/k). Plugging in the values, we get x = √(7 KG)(30 KN/7 KG)/75 KN/m = 0.86 m. Therefore, the distance between the masses when the 4.5 KG mass stops is 0.86 m.

Note: It is always important to check the units and make sure they are consistent throughout the calculations. In this case, we converted the units of the spring constant from KN/m to N/m to match the units of the other quantities.
 
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