91Ó°ÊÓ

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

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.

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

It is given by,

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

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

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

$\begin{array}{rcl}\mathrm{Δ}{U}_g& =& -\mathrm{Δ}{U}_s\\ & =& -0.98\mathrm{J}\end{array}$

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

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

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

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 $3.1\mathrm{N}/\mathrm{cm}$.