From the problem statement, we can calculate how much potential energy is initially stored in the spring. Since we are neglecting air resistance, we can say that all of this energy is lost through friction as the system comes to a stop.

Calculating initial potential energy:

Then we can write an expression for the force of friction on the mass:

We can then write an expression of the work done by friction:

Setting this equal to the initial potential energy:

Therefore, the mass travels a total distance of 20 meters before coming to rest. We know that the distance between maximum compression and maximum extension of the spring is 4 meters. Therefore, we can say that the mass travels from one maximum displacement to the other five times, passing through the equilibrium once each time. The mass passes through the equilibrium point 5 times.

We can calculate the initial potential energy of the spring from what we are given in the problem statement:

Assuming the only energy loss of the spring is through internal friction, we can write:

This problem can be solved by the conservation of energy. The block initially has kinetic energy equal to:

When the velocity of the block is zero, all the kinetic energy has been transferred to potential spring energy.

Since the spring is Hookean, the relationship between the force and displacement from equilibrium of the mass can be expressed by Hooke's Law:

 

Since this equation is linear, the force and displacement are directly proportional. Thus, when the displacement doubles, the force doubles. 

To find the answer to this problem, we'll need to consider the case before the object collides with the spring, and also after it collides. Before the collision, all of the object's energy is in the form of kinetic energy.

Upon collision, the spring will be compressed by the moving object. The maximum displacement, however, will occur the instant the object stops moving. At this point, the object will no longer have any kinetic energy because it has stopped moving. Furthermore, all of the energy has now been transferred into the spring, and is held as spring potential energy.

Also, since we are told in the question stem that no friction is involved, we have a situation in which mechanical energy is conserved. In other words, we can relate the initial and final energies as equal to one another.

Set these two equations equal to each other, and solve for the maximum displacement.

Hooke's Law is:

here  is in meters and  is in newtons. 

Solve for  to see what the units are: 

This means that the units of the spring constant,  is:

We can ignore the negative sign when we determine units, as the negative sign only indicates the direction of the force. 

To determine which type of sinusoid is used, we have to look at its starting point and where it goes. It starts at its minimum and moves towards its maxima. The simplest model we can use is a negative cosine model, which will require no phase shifts. 

At ,  will be at its minimal value. 

However, there is also a counteracting frictional force that decreases the amplitude of the spring over time. One way to do so would be to add an  term in front of the cosine, so that we have an ever decreasing amplitude as time moves on. 

The only appropriate model is

To solve this problem, we must use our knowledge of the relationship between angular frequency, the spring constant, and the mass.

We can then use this, and the derived maximum speed of the object in terms of the amplitude and angular frequency, to find the answer.

Doubling the initial distance the weight is displaced from equilibrium doubles the amplitude, but does not change the period or the phase of the oscillator. So in the general equation of an oscillator:

, only the  term changes, and it doubles.

The amplitude does not change since the student pulls the weight the same displacement. Changing the spring constant changes the frequency of the oscillator. Since the period decreases, the frequency increases, but it does not double. The frequency and the period depend on the square root of the spring constant, so the frequency increases by . In the equation of motion for an oscillator,

, only the frequency , changes.