Question

A small ball with a mass of 60.0 g is loaded into a spring gun,

Answer 1

Answer:

K=70.58 Nw/m

Explanation:

Conservation Of The Energy

When no losses are considered, like friction or air resistance, the total energy present in a system is constant. Three types of energy are commonly used in simple physics applications:

Kinetic energy:

$\displaystyle K=\frac{mv^2}{2}$

Potential gravitational energy:

$U=mgh$

Elastic potential energy

$\displaystyle P=\frac{kx^2}{2}$

The variables used in the formulas are: m=mass, v=speed, h=height above ground, g=acceleration of gravity, k=spring constant, x=spring compression

When the spring is compressed, the ball is at rest, and only two energies are present, the elastic and the potential. Thus

$E_o=P+U$

After the trigger is pressed, all the elastic energy is released and transformed in kinetic and the ball is fired at speed $v_o$. At this moment, the ball has two energies: Kinetic and potential. The potential energy is the same as before because the ball is still at the same height, so

$E_1=K+U$

Equating both energies

$P+U=K+U$

$P=K$

Or equivalently

$\displaystyle \frac{kx^2}{2}=\frac{mv_o^2}{2}$

Rearranging

$kx^2=mv_o^2$

Solving for k

$\displaystyle k=\frac{mv_o^2}{x^2}.......[1]$

We need to find $v_o$ by using motion conditions. We know the ball reaches a distance d=2.2 m after traveling height of h=1.4 m. We can use these data to find $v_o$.

The horizontal distance is given by

$d=v_o.t..........[2]$

The vertical height is

$\displaystyle h=\frac{gt^2}{2}...........[3]$

Solving for t in [2]

$\displaystyle t=\frac{d}{v_o}$

Replacing in [3]

$\displaystyle h=\frac{g\left ( \frac{d}{v_o} \right )^2}{2}$

Operating

$\displaystyle h=\frac{gd^2}{2v_o^2}$

Solving for $v_o^2$

$\displaystyle v_o^2=\frac{gd^2}{2h}$

Replacing in [1]

$\displaystyle k=\frac{m\frac{gd^2}{2h}}{x^2}$

Rearranging

$\displaystyle k=\frac{mgd^2}{2hx^2}$

Computing with m=60 gr= 0.06 Kg, x=12 cm = 0.12 m

$\displaystyle k=\frac{(0.06)(9.8)(2.2)^2}{2(1.4)(0.12)^2}$

$\boxed{K=70.58\ Nw/m}$