Answer:
a) 8.85 kJ
b) 7.85 kJ
c) 6.85 kJ
Explanation:
we will use the energy conservation theorem for all of the stages
[tex]K_1+U_1+W_e=K_2+U_2[/tex]
for stage 1:
the initial velocity and the potential gravitational energy at the bottom is zero:
[tex]0+0+W_e=\frac{1}{2}m*v_f^2+m*g*h\\W_e=\frac{1}{2}*(80kg)(5.00m/s)^2+(80kg)(9.81m/s^2)*(10.0m)\\W_e=8.85kJ[/tex]
for stage 2:
the velocity is constant so there is no change in the kinetic energy, so there is change only on the potential gravitational energy:
[tex]W_e=\Delta U\\W_e=m*g*(h_f-h_i)\\W_e=80kg*9,81m/s^2*(20m-10m)\\W_e=7.85kJ[/tex]
for stage 3:
there is a change for both, gravitational and kinetic energy so:
[tex]W_e=\Delta U + \Delta K\\W_e=m*g*(h_f-h_i)+\frac{1}{2}m*(v_f^2-v_i^2)\\W_e=80*9.81*(30-20)+\frac{1}{2}80*((0)^2-(5.00)^2)\\\\W_e=6.85kJ[/tex]
