A cave rescue team lifts an injured spelunker directly upward and out of a sinkh

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 11.2 m. (1) First, the initially stationary spelunker is accelerated to a speed of 5.20 m/s. (2) He is then lifted at the constant speed of 5.20 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 86.0 kg rescuee by the force lifting him during each stage?

Explanation / Answer

  You want to break this problem up into 3 stages. In each stage you want to find the amount of force that was required. By knowing this force and the distance for each stage (13.0 m) you can find how much work was required for each stage. Finally you can sum up the work from each individual stage to find the total work done during the entire rescue:
STAGE 1:
v^2 = vi^2 + 2*a*?y
5.20^2 = 0^2 + 2*a*11.2
a = 1.21m/s^2

F = ma
F = (86.0 kg) * (1.21 m/s^2)
F = 104.06 N
W1 = F*d
W1 = (104.06 N) * (11.2 m)
W1 = 1165.472 J

STAGE 2:
F = ma
F = 0 (There is a CONSTANT speed. Therefore no acceleration. Since the acceleration is 0, the net force is 0)
W2 = F*d
W2 = (0 N) * (13.0 m)
W2 = 0 J

STAGE 3:
v^2 = vi^2 + 2*a*?y
0^2 = 5.20^2 + 2*a*11.2
a =- 1.21m/s^2

F = ma
F = (86.0 kg) * (1.21 m/s^2)
F = -104.06 N
W1 = F*d
W1 =- (104.06 N) * (11.2 m)
W1 = -1165.472 J

Thus the total work is the work performed during each stage:
W = W1 + W2 + W3
W = 167.57 J + 0 J + (-167.57 J)
W = 0 J

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