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. (1) First, the initially stationary spelunker is accelerated to a speed of 4.30 m/s. (2) Then he is then lifted at the constant speed of 4.30 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 91.0 kg rescuee by the force lifting him during each stage?

Answer 1

Answer:

a) 9.8 kJ

b) 8.9 kJ

c) 8.1 kJ

Explanation:

On the first stage, we have:

$v_i=0 m/s\\v_f=4.30m/s\\h=10m$

applying the energy conservation formula:

$W+K1+U1=K2+U2\\W=K2-K1+U2-U1\\W=\frac{1}{2}*91.0kg*(4.30m/s)^2-(0)+91.0kg*9.8*10-(0)\\\\W=9.8kJ$

for stage 2:

$v_i=4.30 m/s\\v_f=4.30m/s\\hi=10m\\hf=20m$

$W+K1+U1=K2+U2\\W=K2-K1+U2-U1\\W=\frac{1}{2}*91.0kg*(4.30m/s-4.30m/s)^2+91.0kg*9.8*(20-10)\\\\W=8.9kJ$

for the final stage:

$v_i=4.30 m/s\\v_f=0\\hi=20m\\hf=30m$

$W+K1+U1=K2+U2\\W=K2-K1+U2-U1\\W=(0)-\frac{1}{2}*91.0kg*(4.30m/s)^2+91.0kg*9.8*(30-20)\\\\W=8.1kJ$