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:
[tex]\displaystyle K=\frac{mv^2}{2}[/tex]
Potential gravitational energy:
[tex]U=mgh[/tex]
Elastic potential energy
[tex]\displaystyle P=\frac{kx^2}{2}[/tex]
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
[tex]E_o=P+U[/tex]
After the trigger is pressed, all the elastic energy is released and transformed in kinetic and the ball is fired at speed [tex]v_o[/tex]. 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
[tex]E_1=K+U[/tex]
Equating both energies
[tex]P+U=K+U[/tex]
[tex]P=K[/tex]
Or equivalently
[tex]\displaystyle \frac{kx^2}{2}=\frac{mv_o^2}{2}[/tex]
Rearranging
[tex]kx^2=mv_o^2[/tex]
Solving for k
[tex]\displaystyle k=\frac{mv_o^2}{x^2}.......[1][/tex]
We need to find [tex]v_o[/tex] 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 [tex]v_o[/tex].
The horizontal distance is given by
[tex]d=v_o.t..........[2][/tex]
The vertical height is
[tex]\displaystyle h=\frac{gt^2}{2}...........[3][/tex]
Solving for t in [2]
[tex]\displaystyle t=\frac{d}{v_o}[/tex]
Replacing in [3]
[tex]\displaystyle h=\frac{g\left ( \frac{d}{v_o} \right )^2}{2}[/tex]
Operating
[tex]\displaystyle h=\frac{gd^2}{2v_o^2}[/tex]
Solving for [tex]v_o^2[/tex]
[tex]\displaystyle v_o^2=\frac{gd^2}{2h}[/tex]
Replacing in [1]
[tex]\displaystyle k=\frac{m\frac{gd^2}{2h}}{x^2}[/tex]
Rearranging
[tex]\displaystyle k=\frac{mgd^2}{2hx^2}[/tex]
Computing with m=60 gr= 0.06 Kg, x=12 cm = 0.12 m
[tex]\displaystyle k=\frac{(0.06)(9.8)(2.2)^2}{2(1.4)(0.12)^2}[/tex]
[tex]\boxed{K=70.58\ Nw/m}[/tex]
