Exploring the Work Done by a Helicopter on an Astronaut

How much work is done on the astronaut by the force from the helicopter?

Final Answer:

The work done on the astronaut by the force from the helicopter is 1.43 × 10³ J.

Explanation:

The work done on the astronaut by the force from the helicopter can be calculated using the formula:

Work = Force × Distance

Here, the force on the astronaut is equal to their weight, which is given by:

Force = Mass × Acceleration due to Gravity

The distance is the vertical displacement, which is given as 19 m. Plugging in the values:

Work = (82 kg) × (9.8 m/s² / 11) × (19 m)

Work = 1.43 × 10³ J

Therefore, the work done by the force from the helicopter on the astronaut is 1.43 × 10³ J.

When a helicopter lifts an 82 kg astronaut 19 m vertically from the ocean by means of a cable, the concept of work in physics comes into play. Work, in the context of physics, is defined as the product of the force applied to an object and the distance over which the force is exerted. In this scenario, the work done on the astronaut is directly related to the force from the helicopter lifting them and the vertical distance they are raised.

Firstly, to calculate the work done by the force from the helicopter, we need to determine the force acting on the astronaut, which is equivalent to their weight. The weight of an object is given by the formula:

Weight = Mass × Acceleration due to Gravity

Given that the mass of the astronaut is 82 kg and the acceleration of the astronaut is g/11, where g is the acceleration due to gravity, we can calculate the force acting on the astronaut. Then, the work done can be found using the formula for work:

Work = Force × Distance

Substitute the values of force (weight of the astronaut) and the vertical distance of 19 m into the formula to get the final answer, which is 1.43 × 10³ J. This indicates the amount of work done on the astronaut by the force from the helicopter during the vertical lift.

← Arrival time challenge museum visit How to calculate the resistance of a 5 ampere fuse wire in a circuit →