Question

A $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{g}$marble is fired vertically upward using a spring gun. The spring must be compressed $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$if the marble is to just reach a target $\mathbf{20}\mathbf{}\mathbf{m}$above the marble’s position on the compressed spring.

(a) What is the change $\mathit{\Delta }{\mathit{U}}_{\mathbf{g}}$in the gravitational potential energy of the marble-Earth system during the $\mathbf{20}\mathbf{}\mathbf{m}$ ascent?

(b) What is the change $\mathit{\Delta }{\mathit{U}}_{\mathbf{s}}$in the elastic potential energy of the spring during its launch of the marble?

(c) What is the spring constant of the spring?

Short answer

(a) Change in gravitational potential energy is$0.98\hspace{0.17em}\text{J}$ .

(b) Change in elastic potential energy of the spring during launch of marble is $-0.98\mathrm{J}$.

(c) Spring constant of spring is $3.1\mathrm{N}/\mathrm{cm}$.

Step-by-step solution

Step 1: Given data

1. Mass of marble is $M=5.0\times {10}^{-3}\mathrm{kg}$,
2. Compression of spring is $x=8.0\text{cm} \text{or} 0.0\text{8m}$,

Determining the concept

The problem deals with the law of conservation of energywhich states that the total energy of an isolated system remains constant. Using energy conservation, the change in gravitational potential energy, as well as the change in elastic energy and spring constant of spring, can be found.

Formulae are as follows:

Elastic potential energy is given by,

$E=\frac{1}{2}k{x}^2$ ...(i)

$PE=mgh$ ...(ii)

where m is mass, k is spring constant, x is displacement, g is the acceleration due to gravity, h is height, PE is potential energy andE is kinetic energy.

(a) Determining the change in gravitational potential energy

At the top point, the marble has potential energy, thus using the notation in equation (ii) gravitational potential energy.

It is given by,

$\begin{array}{rcl}{U}_g& =& mgh\\ & =& \left(5.0\times {10}^{-3}\right)\left(9.8\right)\left(20\right)\\ & =& 0.98\text{J}\end{array}$

Hence, the change in gravitational potential energy is $0.98\mathrm{J}$.

(b) Determining the change in elastic potential energy of the spring during launch of marble

As the kinetic energy is zero along the top and launch points. So, according to energy conservation, it can be written as,

$\begin{array}{rcl}\Delta U_g& =& -\Delta U_s\\ & =& -0.98\mathrm{J}\end{array}$

Hence, thechange in elastic potential energy of the spring during launch of marble is$-0.98\mathrm{J}$

(c) Determining the spring constant of spring

As, the total energy of the system remains constant, it can be written as,

$\begin{array}{rcl}E& =& \frac{1}{2}\times k{x}^2\\ k& =& \frac{2E}{x^2}\end{array}$

As, the results in parts a & b the energy of system is $0.98\mathrm{J}$,

$\begin{array}{rcl}k& =& \frac{2\left(0.980\right)}{{0.080}^2}\\ & =& 306.25\mathrm{N}/\mathrm{m}\end{array}$

role="math" $k=3.1\mathrm{N}/\mathrm{cm}$

Hence, the spring constant of spring is $3.1\mathrm{N}/\mathrm{cm}$.