Question

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) 200 j

B) 0 J

C) -100 J

D) -300 J

Step-by-step explanation:

We know that change in potential energy is given as

`\Delta U =  U_f - U_i = mgy_f - mgy_i`

where, mg is weight

yf and yi final and initial position

mg = 10 N

yf = 0 m

yi = 20 m

`\Delta U = 10* 20 - 0 = 200 J`

B) mg = 10 N

yf = 20 m

yi = 20 m

`\Delta U =  0 j`

c) before fall from the drone

U = mgh

h = yf - yi = 20 - 30 = -10 m

`U = 10* (-10) = -100 J`

D) point has 0 potential energy

mg = 10 N

h = yf -yi = 0 -30 = - 30 m

`U = 10 * (-30) = -300 J`