Answer:
K=70.58 Nw/m
Step-by-step 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=(mv^2)/(2)](https://img.qammunity.org/2021/formulas/physics/middle-school/b0tvh6ctnujy288f9a6om8zum6nb3izon6.png)
Potential gravitational energy:
![U=mgh](https://img.qammunity.org/2021/formulas/physics/middle-school/g40n03woz39gur7y23rmsdjh1j6xykiiwr.png)
Elastic potential energy
![\displaystyle P=(kx^2)/(2)](https://img.qammunity.org/2021/formulas/physics/college/tzr2273xcklu348yfdl8gcb7tdilq0s820.png)
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](https://img.qammunity.org/2021/formulas/physics/college/53ptrdwx0hyzqlsvmgapvtqqnr3d71bq5h.png)
After the trigger is pressed, all the elastic energy is released and transformed in kinetic and the ball is fired at speed
. 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](https://img.qammunity.org/2021/formulas/physics/college/lc3muy3yvrko1xwuqg4jmb0di96gvrn445.png)
Equating both energies
![P+U=K+U](https://img.qammunity.org/2021/formulas/physics/college/i7lhfbftg893rtf2ez4ix125o7t5w1rdy5.png)
![P=K](https://img.qammunity.org/2021/formulas/physics/college/cp1rfvm02kka63rj5wvxyg63mvhs9fxqd3.png)
Or equivalently
![\displaystyle (kx^2)/(2)=(mv_o^2)/(2)](https://img.qammunity.org/2021/formulas/physics/college/xy186tfql53i9uwj14pyx0ymbbokfi6mfk.png)
Rearranging
![kx^2=mv_o^2](https://img.qammunity.org/2021/formulas/physics/college/ge7o71tpudbtu1eeuobq942ytv10l5zb4j.png)
Solving for k
![\displaystyle k=(mv_o^2)/(x^2).......[1]](https://img.qammunity.org/2021/formulas/physics/college/ofzyzwtf1nqnesz4l5t8wlzdk3vx8vc2pc.png)
We need to find
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
.
The horizontal distance is given by
![d=v_o.t..........[2]](https://img.qammunity.org/2021/formulas/physics/college/cby5xz70wcn7p939oiaeq9wuyovv9lhnh6.png)
The vertical height is
![\displaystyle h=(gt^2)/(2)...........[3]](https://img.qammunity.org/2021/formulas/physics/college/ni2yruu95f1io5up9dxko40ms2r0d0wmjn.png)
Solving for t in [2]
![\displaystyle t=(d)/(v_o)](https://img.qammunity.org/2021/formulas/physics/college/y4lh434jjyi9djnowlkxdvgfix7mdwabfl.png)
Replacing in [3]
![\displaystyle h=(g\left ( (d)/(v_o) \right )^2)/(2)](https://img.qammunity.org/2021/formulas/physics/college/50qpjrja3h1c7zsjf788k1x7i8822ki888.png)
Operating
![\displaystyle h=(gd^2)/(2v_o^2)](https://img.qammunity.org/2021/formulas/physics/college/vcd65dmg7cy038cnosrgbno3czdnmbpfuj.png)
Solving for
![v_o^2](https://img.qammunity.org/2021/formulas/physics/college/ii8934upiml3l6u7o6q8pnakc8i2rywdrf.png)
![\displaystyle v_o^2=(gd^2)/(2h)](https://img.qammunity.org/2021/formulas/physics/college/k12yhgwliapjsawd4td33if28zqyh6y8t1.png)
Replacing in [1]
![\displaystyle k=(m(gd^2)/(2h))/(x^2)](https://img.qammunity.org/2021/formulas/physics/college/v49ikuamcorhw37l3vdel06k7i2ot1anm3.png)
Rearranging
![\displaystyle k=(mgd^2)/(2hx^2)](https://img.qammunity.org/2021/formulas/physics/college/rogephfgzf7zjsx3ds891qur652p0i1ydn.png)
Computing with m=60 gr= 0.06 Kg, x=12 cm = 0.12 m
![\displaystyle k=((0.06)(9.8)(2.2)^2)/(2(1.4)(0.12)^2)](https://img.qammunity.org/2021/formulas/physics/college/whyv3rta6mksuby25lw3ovg20vtbhcw2tm.png)
![\boxed{K=70.58\ Nw/m}](https://img.qammunity.org/2021/formulas/physics/college/gikioi99i5boffo95vv1oc6zuw8cqadl1q.png)