Asked by joy
A helicopter rescues a trapped person of mass m = 67.0 kg
from a flooded river by lifting the person vertically upward using a winch and rope. The person is pulled 13.0 m into the helicopter with a constant force that is 11% greater than the person's weight.
(a) Find the work done by each of the forces acting on the person. (Enter your answers to the nearest whole number. Be sure to use 9.81 m/s2 for g.)
Wg =J
WT = J
(b) Assuming the survivor starts from rest, determine his speed upon reaching the helicopter.
m/s
from a flooded river by lifting the person vertically upward using a winch and rope. The person is pulled 13.0 m into the helicopter with a constant force that is 11% greater than the person's weight.
(a) Find the work done by each of the forces acting on the person. (Enter your answers to the nearest whole number. Be sure to use 9.81 m/s2 for g.)
Wg =J
WT = J
(b) Assuming the survivor starts from rest, determine his speed upon reaching the helicopter.
m/s
Answers
Answered by
bobpursley
assuming no friction,
work=force*distance=1.13*mg*13
b. if the excess work went into KEnergy, then
kinetic energy=.13*mg*13.0
then KE= 1/2 m v^2 and you can solve for v.
work=force*distance=1.13*mg*13
b. if the excess work went into KEnergy, then
kinetic energy=.13*mg*13.0
then KE= 1/2 m v^2 and you can solve for v.
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