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 4.60 m/s. (2) He is then lifted at the constant speed of 4.60 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 74.0 kg rescuee by the force lifting him during each stage?

Answer 1

For the 1st stage,

the distance traveled, s is given by :

v^2 = u^2 + 2as

where u = initial velocity = 0

v = 4.6 m/s

s = 11.2

So, 4.6^2 = 0 + 2*a*11.2

So, a = 0.945 m/s2

So, the force duting this time , F = m*a = 74*0.945 = 69.9 N

So, the Work done = F*s = 69.9*11.2 = 782.9 J

For the 2nd stage,

the acceleration is 0 as the speed is constant

So, Force is 0

So, work done = 0

For the third stage,

As the final Kinetic energy of the body is 0,

So, negative work must be done so that the body comes to rest.

So, Work done in this stage = -782.9 J