Answer:
a. [tex]0.4415m/s[/tex]
b. 0.5530m/s
Explanation:
a. Mass of the fish is given as [tex]m=2.85kg[/tex] and the spring's force constant as [tex]k=895N/m[/tex]. We take [tex]y=0[/tex] as our initial position hence:[tex]y_1=0,\ x_1=0, \ v_1=0[/tex]
Applying the work-energy theorem:
[tex]K_1+U_g_1+U_e_l_1+W_o_t_h_e_r=K_2+U_g_2+U_e_l_2 \ \ \ # No\ other \ forces \ hence \ W_o_t_h_e_r=0\\\\ \\K_1+U_g_1+U_e_l_1=K_2+U_g_2+U_e_l_2 \\K=0.5mv^2\\U_g=mgy\\U_e_l=0.5kx^2\\[/tex]
#Take point 2, [tex]0.05m[/tex] below initial position.
[tex]y_2=-0.05m,\ x_2=0.05, \ v_2=0[/tex]
Since the fish starts at rest with zero gravitational potential energy, we obtain: [tex]K_1=0, \ U_g_1=0[/tex], and [tex]U_e_l_1=0[/tex] for initially either stretched/compressed spring.
Therefore:
[tex]U_g_2=2.85\times 9.8\times-0.05\\=-1.3965J[/tex]
And plug values for x2 and k into [tex]U_e_l_2[/tex] equation:
[tex]U_e_l_2=0.5\times 895 \times 0.05^2\\=1.11875J[/tex]
Substituting energy quantities into [tex]K_1+U_g_1+U_e_l_1=K_2+U_g_2+U_e_l_2[/tex] to get:
[tex]0+0+0=K_2-1.3965+1.11875\\K_2=0.27775J[/tex]
#Substitute in [tex]K=0.5mv^2[/tex] :
[tex]0.27775=0.5(2.85)v_2^2\\v_2=0.4415m/s[/tex]
b.To find the maximum speed of the fish as it descends:
#Maximum speeds occurs when kinetic energy is maximum (at what point the slope is zero)
[tex]\frac{dK_2}{dy}=mg-ky=0\\ky=mg\\y=\frac{mg}{k}\\[/tex]#Substitute values for m and K
[tex]y=\frac{2.85\times 9.8}{895} \ \ \g=9.8\\y=0.03121m[/tex]
Hence:
[tex]K_2=2.85\times 9.8\times 0.03121-0.5\times 895\times 0.03121^2\\=0.4358J[/tex]
#Plug our [tex]K_2[/tex] value in [tex]K=0.5mv^2[/tex] to obtain [tex]v_m_a_x[/tex]
[tex]0.4358=0.5(2.85)v_m_a_x^2\\\\v_m_a_x=0.5530m/s[/tex]
Hence the maximum speed of the fish during descent is 0.5530m/s
