Answer:
22m/s
Explanation:
Mass, m=60 kg
Force constant, k=1300N/m
Restoring force, Fx=6500 N
Average friction force, f=50 N
Length of barrel, l=5m
y=2.5 m
Initial velocity, u=0
[tex]F_x=kx[/tex]
Substitute the values
[tex]6500=1300x[/tex]
[tex]x=\frac{6500}{1300}=5[/tex]m
Work done due to friction force
[tex]W_f=fscos\theta[/tex]
We have [tex]\theta=180^{\circ}[/tex]
Substitute the values
[tex]W_f=50\times 5cos180^{\circ}[/tex]
[tex]W_f=-250J[/tex]
Initial kinetic energy, Ki=0
Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\
Initial elastic potential energy
[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]
[tex]U_{el,1}=16250J[/tex]
Final elastic energy,[tex]U_{el,2}=0[/tex]
Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]
Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]
Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]
Using work-energy theorem
[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]
Substitute the values
[tex]0+0+16250-250=30v^2+1470+0[/tex]
[tex]16000-1470=30v^2[/tex]
[tex]14530=30v^2[/tex]
[tex]v^2=\frac{14530}{30}[/tex]
[tex]v=\sqrt{\frac{14530}{30}}[/tex]
[tex]v=22m/s[/tex]
