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

User Cancerian
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Final answer:

The work done on the spelunker by the lifting force is 1000 J for acceleration, 7840 J for the constant-speed lift, and -1000 J for deceleration.

Step-by-step explanation:

Stage (a): Acceleration

To accelerate the spelunker from 0 m/s to 5.00 m/s, we can calculate the work done using the work-energy principle, which states that the work done is equal to the change in kinetic energy.

The initial kinetic energy is 0 because the spelunker is stationary, and the final kinetic energy is given by KE = 1/2mv². Substituting the values, we have:

Work = ∆KE = 1/2(80.0 kg)(5.00 m/s)²

= 1000 J

Stage (b): Constant Speed

When the spelunker is lifted at a constant speed, the only force doing work is the lifting force against gravity.

Since the speed is constant, the net work done is simply the work done to overcome the gravitational force, which is equal to the change in gravitational potential energy. Using GPE, the work done is:

Work = GPE = (80.0 kg)(9.8 m/s²)(10.0 m)

= 7840 J

Stage (c): Deceleration

Decelerating the spelunker to zero speed involves doing negative work on the spelunker’s kinetic energy, reducing it back to zero. Thus, the work done during deceleration is the negative of the work done during acceleration, which is:

Work = -1000 J

Note that the negative work done by the force during deceleration indicates that the force is taking energy out of the system, hence slowing down the spelunker.

User Max Izrin
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