Question

A solid block, with a mass of 8.58kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring constant value of 368N/m, until the spring is compressed by 11cm. The block is then released. What is the block's final speed, in m/s?

Answer 1

In this problem, we are talking about Mechanical Energy ($M$) which is the addition of the Kinetic Energy $K$ (energy of the body in motion) and Potential Energy $P$ (It can be Gravitational Potential Energy or Elastic Potential Energy, in this case is the second):

$M=K+P$   (1)

The Kinetic Energy is: $K=\frac{1}{2}mV^{2}$

Where $m$ is the mass of the body and $V$ its velocity

And the Potential Energy (Elastic) is: $P=\frac{1}{2}cX^{2}$

Where $C$ is the spring constant and $X$ is the the position of the body.

Knowing this, the equation for the Mechanical Energy in this case is:

$M=\frac{1}{2}mV^{2}+\frac{1}{2}cX^{2}$   (2)

Now, according to the Conservation of the Energy Principle, and knowing there is not friction, the initial energy $M_{i}$ must be equal to the final energy $M_{f}$:

$M_{i}=M_{f}$    (3)

$M_{i}=\frac{1}{2}m{V_{i}}^{2}+\frac{1}{2}c{X_{i}^{2}}$    (4)

At the beginning, the block has a $V_{i}=0$, because it starts from rest, this means the initial energy $M_{i}$ is only the Potential Elastic Energy:

$M_{i}=\frac{1}{2}c{X_{i}^{2}}$    (5)

After the spring is compressed, is in its equilibrium point X=0, so $X_{f}=0$. Then the block is released. This means the final energy $M_{f}$ is only the Kinetic Energy

$M_{f}=\frac{1}{2}m{V_{f}}^{2}$    (6)

Now, we have to substitute (5) and (6) in (3):

$\frac{1}{2}c{X_{i}^{2}}=\frac{1}{2}m{V_{f}}^{2}$

$V_{f}=\sqrt{\frac{c{X_{i}^{2}}}{m}}$

$V_{f}= X_{i}\sqrt{\frac{c}{m}}$

$V_{f}=0.11m\sqrt{\frac{368N/m}{8.58kg}}$

Finally:

$V_{f}=0.7203m/s$