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 1s performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spe- lunker 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 decel- erated to zero speed. how much work is done on the 80.0 kg rescuee by the force lifting him during each stage

Answer 1

Below

Step-by-step explanation:

Work = force x distance ==> the forces during the 3 stages are different

a) I will assume accel is constant for the first 10 m

Average velocity = 5/2 = 2.5 m/s

10 m / 2.5 m/s = 4 seconds

then a = 5 m/s / 4 s = 1.25 m/s^2 <==this accel is added to 'g'

F = ma

= 80 * (9.81 + 1.25) = 884.8 N

W = 884.8 * 10 = 8848 J

b) This one is easier W = F * d = (80)(9.81)(10) = 7848 J

c) Now we decelerate at 1.25 m/s^2 <==== this SUBTRACTS from g

W = ( 80 )(9.81-1.25)*10 = 6848 J

Answer 2

C

Step-by-step explanation:

Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed. C