Question

21. A camera weighing 10 N falls from a small drone hovering 20 m overhead and enters free fall. What is the gravitational potential energy change of the camera from the drone to the ground if you take a reference point of (a) the ground being zero gravitational potential energy? (b) The drone being zero gravitational potential energy? What is the gravitational potential energy of the camera (c) before it falls from the drone and (d) after the camera lands on the ground if the reference point of zero gravitational potential energy is taken to be a second person looking out of a building 30 m from the ground?

Answer 1

a) -200J

b) -200J

c) -100J

d) -300J

Step-by-step explanation:

We know that:

Gravitational potential energy= P.E = mgh

Change in potential energy = Final potential energy – Initial potential energy

∆P.E=P.E2-P.E1

a) When ground being zero potential energy point:

Height of drone with respect to reference point = 20m

Height of Ground with respect to reference point = 0

Potential energy of camera at the drone= P.E1 = mgh = (10)(20) = 200J

Potential energy of camera at the ground= P.E2 = 0

∆P.E=0-200 = - 200J

b) When drone being zero potential energy point:

Height of drone with respect to reference point = 0

Height of Ground with respect to reference point = -20m

Potential energy of camera at the drone= P.E1 = 0

Potential energy of camera at the ground= P.E2 = mgh = (10)(-20) = -200J

∆P.E=-200-0 = - 200J

c) Reference point is a person looking out of window 30m above the ground:

Height of drone with respect to reference point = -10m

Potential energy of camera at the drone= P.E2 = (10)(-10) = -100

Potential energy of camera at reference point = P.E1 = 0

∆P.E=-100-0 = - 100J

d) Reference point is a person looking out of window 30m above the ground:

Height of ground with respect to reference point = -30m

Potential energy of camera at the ground= P.E2 = (10)(-30) = -300

Potential energy of camera at reference point = P.E1 = 0

∆P.E=-300-0 = - 300J