A helicopter lifts a 70 kg astronaut 13 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10.
(a) How much work is done on the astronaut by the force from the helicopter? J
(b) How much work is done on the astronaut by her weight? J
(c)What is the kinetic energy? J
(d) What is the speed of the astronaut just before she reaches the helicopter? m/s
Start with calculating the final velocity v (upwards) of the astronaut (m=70 kg) who starts from rest (v0=0) subject to an acceleration of a=0.1g and through a distance of H=13m.
v²-v0²=2aH
solve for v.
a. sum potential and kinetic energies
potential energy gain, Ep = mgH
kinetic energy gain, Ek = (1/2)mv²
Work done = Ep + Ek
b. Ep
c. Ek
d. v
To solve this problem, we need to use the principles of work, energy, and acceleration. Let's break it down step by step.
(a) Work done by the helicopter's force:
The work done on an object can be calculated using the formula:
Work = Force × Distance × cos(θ)
In this case, the force is equal to the weight of the astronaut, which is given by:
Force = mass × acceleration due to gravity = 70 kg × 9.8 m/s² = 686 N
The distance moved vertically is given as 13 m. Since the force and the displacement are in the same direction, the angle between them is 0 degrees, and the cosine of 0 degrees is 1.
Therefore, the work done by the helicopter's force is:
Work = 686 N × 13 m × 1 = 8922 J
(b) Work done by the astronaut's weight:
The work done by the weight force is given by the same formula as (a), where the force is the component of the weight force acting along the displacement. In this case, the angle between the weight force and the vertical displacement is 180 degrees, and the cosine of 180 degrees is -1.
Therefore, the work done by the astronaut's weight is:
Work = 686 N × 13 m × -1 = -8922 J
(c) Kinetic energy:
Before reaching the helicopter, the astronaut has done work against her weight. According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy.
Since the astronaut started from rest, her initial kinetic energy is zero. The work done against her weight is equal to the final kinetic energy.
Therefore, the kinetic energy is:
Kinetic Energy = Work done against weight = 8922 J
(d) Speed of the astronaut just before reaching the helicopter:
The kinetic energy formula can be used to calculate the speed of an object. The formula is:
Kinetic Energy = (1/2) × mass × velocity²
Rearranging the formula to solve for velocity gives us:
Velocity = √(2 × Kinetic Energy / mass)
Plugging in the values, we get:
Velocity = √(2 × 8922 J / 70 kg) ≈ 19.3 m/s
So, the speed of the astronaut just before she reaches the helicopter is approximately 19.3 m/s.