Respuesta :
Answer:
Explanation:
Given
at certain point displacement and velocity is 0.345 m and 0.54 m/s
Therefore Potential and Kinetic Energy associated is
[tex]U=\frac{1}{2}kx^2[/tex]
[tex]U=\frac{1}{2}\times 9.8\times (0.345)^2[/tex]
[tex]U=0.583 J[/tex]
Kinetic Energy [tex]K=\frac{1}{2}mv^2[/tex]
[tex]k=\frac{1}{2}\times 1.2\times (0.54)^2=0.174 J[/tex]
Total Energy [tex]T=U+K=0.583+0.174=0.757 J[/tex]
Total Energy is conserved hence at maximum displacement all energy will be Potential energy
[tex]T=\frac{1}{2}kA^2[/tex]
where A=maximum displacement
[tex]0.757=\frac{1}{2}\times 9.8\times A^2[/tex]
[tex]0.1544=A^2[/tex]
[tex]A=0.393 m[/tex]
Maximum speed occurs at equilibrium Position where Potential Energy is zero
thus [tex]T=\frac{1}{2}mv^2[/tex]
[tex]0.757=\frac{1}{2}\times 1.2\times v_{max}^2[/tex]
[tex]v_{max}=1.123 m/s[/tex]
(c)When block is at 0.2 m from Equilibrium speed then its Potential Energy is
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}\times 9.8\times (0.2)^2=0.196 J[/tex]
[tex]T=U+K[/tex]
[tex]K=0.757-0.196=0.561 J[/tex]
[tex]K=\frac{1}{2}mv^2[/tex]
[tex]0.561=\frac{1}{2}\times 1.2\times v^2[/tex]
[tex]v=0.966 m/s[/tex]
