Finding the final distance for the compressing in a spring system

In summary, the problem involves a box of mass 5.00 kg sliding on a horizontal frictionless surface with an initial speed of 2.40 m/s. The box encounters a spring with a constant of 2230 N/m and comes to a temporary stop after compressing the spring some distance. The question asks if the Work-Energy Theorem can be used to solve the problem and provides step-by-step directions. The solution involves using Newton's 2nd law and Hooke's law to write an ODE, solving for the initial conditions, and using energy considerations to find the final compression distance of the spring. The final formula for the compression distance is x_c = √(m/k) * v_i.
  • #1
cbarker1
Gold Member
MHB
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Dear Every one,

Here is the question to the problem:

As shown in the figure below, a box of mass
m = 5.00 kg is sliding across a horizontal frictionless surface with an initial speed vi = 2.40 m/s when it encounters a spring of constant k = 2230 N/m. The box comes momentarily to rest after compressing the spring some amount xc. Determine the final compression xc of the spring.
7-p-028.gif



Is there way to use the Work-Energy Thereom to solve this problem? If so, you help me through step by step directions?
Thanks,
Cbarker1
 
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  • #2
kinetic energy of the mass = potential energy stored in the spring

graph of kinetic energy and elastic potential energy vs position ...
 
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  • #3
Let's look at this dynamically, where time $t=0$ coincides with the time the box first contacts the spring, and orient our $x$-axis of motion such that the origin coincides with the point where the box first contacts the spring. Using Newton's 2nd law of motion, and Hooke's law we obtain the IVP:

\(\displaystyle mx''(t)=-kx(t)\) where \(\displaystyle x'(0)=v_i,\,x(0)=0\)

Let's write the ODE in standard linear form:

\(\displaystyle x''(t)+\frac{k}{m}x(t)=0\)

By the theory of 2nd order linear homogeneous ODE's, we know the solution will take the form:

\(\displaystyle x(t)=c_1\cos\left(\sqrt{\frac{k}{m}}t\right)+c_2\sin\left(\sqrt{\frac{k}{m}}t\right)\)

Hence:

\(\displaystyle x'(t)=-c_1\sqrt{\frac{k}{m}}\sin\left(\sqrt{\frac{k}{m}}t\right)+c_2\sqrt{\frac{k}{m}}\cos\left(\sqrt{\frac{k}{m}}t\right)\)

Using the given initial conditions, we obtain the system:

\(\displaystyle 0=c_1\)

\(\displaystyle v_i=c_2\sqrt{\frac{k}{m}}\implies c_2=\sqrt{\frac{m}{k}}v_i\)

And so the solution to the IVP is given by:

\(\displaystyle x(t)=\sqrt{\frac{m}{k}}v_i\sin\left(\sqrt{\frac{k}{m}}t\right)\) where \(\displaystyle 0\le t\le\sqrt{\frac{m}{k}}\pi\)

And we have:

\(\displaystyle x'(t)=v_i\cos\left(\sqrt{\frac{k}{m}}t\right)\)

When the spring reaches maximal compression, we must have:

\(\displaystyle x'(t)=0\implies \sqrt{\frac{k}{m}}t=\frac{\pi}{2}\implies t=\sqrt{\frac{m}{k}}\cdot\frac{\pi}{2}\)

And so we find:

\(\displaystyle x_c=x\left(\sqrt{\frac{m}{k}}\cdot\frac{\pi}{2}\right)=\sqrt{\frac{m}{k}}v_i\sin\left(\sqrt{\frac{k}{m}}\cdot\sqrt{\frac{m}{k}}\cdot\frac{\pi}{2}\right)=\sqrt{\frac{m}{k}}v_i\)

Now, if we use energy considerations, we know the initial kinetic energy of the box must equal the work done to compress the spring when the spring reaches maximal compression:

\(\displaystyle \frac{1}{2}mv_i^2=\frac{1}{2}kx_c^2\)

Solving for $x_c$, we find:

\(\displaystyle x_c=\sqrt{\frac{m}{k}}v_i\quad\checkmark\)

Notice this formula is not only dimensionally consistent, but it also makes sense when we look at what happens when we change any of the parameters:

  • Increase/decrease the mass of the box, and the compression distance increases/decreases too.
  • Increase/decrease the stiffness of the spring and the compression distance decreases/increases.
  • Increase/decrease the initial velocity of the box, and the compression distance increases/decreases too.
 

FAQ: Finding the final distance for the compressing in a spring system

What is the formula for finding the final distance in a spring system?

The formula for finding the final distance in a spring system is d = F/k, where d represents the final distance, F represents the applied force, and k represents the spring constant.

How do you determine the spring constant in a spring system?

The spring constant can be determined by dividing the applied force by the displacement of the spring. This can be represented as k = F/d, where k is the spring constant, F is the applied force, and d is the displacement.

What is the unit of measurement for the spring constant?

The unit of measurement for the spring constant is N/m, which stands for Newtons per meter.

How does the final distance change if the applied force is doubled?

If the applied force is doubled, the final distance will also double. This is because the final distance is directly proportional to the applied force in a spring system.

Can the final distance be negative in a spring system?

Yes, the final distance can be negative in a spring system. This means that the spring has been compressed and the particles are closer together than their equilibrium position.

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