The work done by air resistance during the descent from the height of the helicopter to the ground is [tex]1.32 \times 10^5 \ J[/tex].
The given parameters:
The work done by air resistance during the descent from the height of the helicopter to the ground is calculated by applying work-energy theorem as follows;
W = ΔE
[tex]W = P.E - K.E\\\\W = mgh \ - \ \frac{1}{2} mv^2\\\\W = (15 \times 9.8 \times 1000) -(\frac{1}{2} \times 15 \times 45^2)\\\\W = 1.32 \times 10^5 \ J[/tex]
Thus, the work done by air resistance during the descent from the height of the helicopter to the ground is [tex]1.32 \times 10^5 \ J[/tex].
The complete question is below:
A helicopter is hovering at an altitude of 1 km when a panel from its underside breaks loose and plummets to the ground. The mass of the panel is 15 kg and it hits the ground with speed of 45 m/s. Calculate the work done by air resistance during the descent from the height of the helicopter to the ground.
Learn more about work-energy theorem here: https://brainly.com/question/22236101
