Question

in an arcade game a 0.14 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releasing it. if the spring has a spring constant of 155 n/m and is compressed from its equilibrium position by 4 cm, find the speed with which the disk slides across the surface. answer in units of m/s.

Answer 1

Answer:

Approximately $1.3\; {\rm m\cdot s^{-1}}$.

Explanation:

When an ideal spring with spring constant $k$ is displaced by $x$ from equilibrium, the elastic potential energy (${\rm EPE}$) stored in the spring will be:

$\begin{aligned}{\rm PE} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$.

If an object of mass $m$ is travelling at a speed of $v$, the kinetic energy (${\rm KE}$) of that object will be:

$\begin{aligned}{\rm KE} &= \frac{1}{2}\, m\, v^{2}\end{aligned}$.

Under the assumptions of this question, all the elastic potential energy (${\rm EPE}$) stored in the spring would have been turned into the kinetic energy (${\rm KE}$) of the disk.

Let $k$ and $x$ denote the spring constant and displacement from equilibrium of this spring. Let $m$ and $v$ denote the mass and speed of the disk. Apply unit conversion and ensure that the displacement of the spring is in standard units (meters): $x = 4\; {\rm cm} = 0.04\; {\rm m}$.

${\rm EPE} = {\rm KE}$.

$\begin{aligned}\frac{1}{2}\, m\, v^{2} &= {\rm KE} = {\rm EPE} = \frac{1}{2}\, k\, x^{2}\end{aligned}$.

Rearrange this equation to find the speed of the disk $v$:

$\begin{aligned}v^{2} = \frac{k\, x^{2}}{m}\end{aligned}$.

Since speed $v \ge 0$:

$\begin{aligned}v &= \sqrt{\frac{k\, x^{2}}{m}} \\ &= x\, \sqrt{\frac{k}{m}}\\ &= 0.04\; {\rm m} \times\sqrt{\frac{155\; {\rm N \cdot m^{-1}}}{0.14\; {\rm kg}}} \\ &\approx 1.3\; {\rm m\cdot s^{-1}}\end{aligned}$.

(Note that $1\; {\rm N \cdot m^{-1}} = 1\; {\rm (kg \cdot m\cdot s^{-2}) \cdot m^{-1}} = 1\; {\rm kg \cdot s^{-2}}$.)

Hence, the speed of the disk would be approximately $1.3\; {\rm m\cdot s^{-1}}$.