A rescue helicopter lifts a 65 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 13 m.
(a) What is the tension in the cable?
N
(b) How much work is done by the tension in the cable?
J
(c) How much work is done by the person's weight?
J
(d) Use the work-energy theorem and find the final speed of the person.
m/s
4 answers
What is your thinking?
(a) Start by drawing a free body diagram for the person and applying Newton's second law, F = m a. The "F" will be the difference between the cable tension T and the weight M g. Solve for T.
(b) T*(distance moved) = work done by cable
(c) -M*g x (distance moved) = work done by person's weight
(d) Total work done = (b) + (c) = change in kinetic energy
(b) T*(distance moved) = work done by cable
(c) -M*g x (distance moved) = work done by person's weight
(d) Total work done = (b) + (c) = change in kinetic energy
how do you solve for the final speed?
vf= sqrt( (2 Kef)/m)
0 answers