Answer:11.18 m/s
Explanation:
Given
mass of particle m=0.2 kg
Potential Energy U(x) is given by
[tex]U(x)=8x^2+2x^4[/tex]
at x=1 m
[tex]U(1)=8+2=10 J[/tex]
kinetic energy at x=1 m
[tex]K.E.=\frac{1}{2}mv^2=\frac{1}{2}\times 0.2\times 5^2[/tex]
[tex]K.E.=2.5 J[/tex]
Total Energy =U+K.E.
[tex]Total=10+2.5=12.5 J[/tex]
at x=0, U(0)=0
as total Energy is conserved therefore K.E. at x=0 is equal to Total Energy
[tex]\frac{1}{2}\times 0.2\times v^2=12.5[/tex]
[tex]v^2=125[/tex]
[tex]v=\sqrt{125}[/tex]
[tex]v=11.18 m/s[/tex]
