Question

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable.The lift is performed in three stages, each requiring a vertical distance of 10.0 m:

(a) the initially stationary spelunker is accelerated to a speed of 5.00m/s;
(b) he is then lifted at the constant speed of 5.00m/s;
(c) finally he is decelerated to zero speed. How much work is done on the 80.0kg rescuee by the force lifting him during each stage?

Answer 1

a) 8.85 kJ

b) 7.85 kJ

c) 6.85 kJ

Step-by-step explanation:

we will use the energy conservation theorem for all of the stages

`K_1+U_1+W_e=K_2+U_2`

for stage 1:

the initial velocity and the potential gravitational energy at the bottom is zero:

`0+0+W_e=(1)/(2)m*v_f^2+m*g*h\\W_e=(1)/(2)*(80kg)(5.00m/s)^2+(80kg)(9.81m/s^2)*(10.0m)\\W_e=8.85kJ`

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:

`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`

for stage 3:

there is a change for both, gravitational and kinetic energy so:

`W_e=\Delta U + \Delta K\\W_e=m*g*(h_f-h_i)+(1)/(2)m*(v_f^2-v_i^2)\\W_e=80*9.81*(30-20)+(1)/(2)80*((0)^2-(5.00)^2)\\\\W_e=6.85kJ`