Question

A 10 kg rock is released from a height of 27 m. In the following answer boxes, enter the total mechanical energy of the rock at each of the following heights: 13.50 m, 9.00 m, 6.75 m. Neglect air-resistance in your calculations.

Answer 1

Okay, here's the solution~

i) At 27 m of height,
kinetic energy (KE) is given by,

KE = 1/2 (m)(v)^2
where, m = mass = 10 kg
v = velocity at 27m height = 0 m/s (since it is released from rest)

therefore KE = 0 J

potential energy(PE) is given by

PE = mgh

where, m = mass = 10kg
g = acceleration due to gravity = 10 m/s^2
h = height = 27m

PE = 27x10x10

PE = 2700 J

total energy is given by,

TE = KE + PE

TE = 0 + 2700

TE = 2700 J

Now, at any point if we check during the flight of the stone, the energy remains conserved.

So, at any point (be it 13.50m, 9.00m 6.75m ), the total mechanical energy remains same that is 2700J.

However, I believe this is not your question and you mean to ask the independant PE and KE at each height.

Therefore, solving one for you.

ii) Height = 13.50m

Total Energy remains conserved (i.e. unchanged) throughout the flight, therefore

TE = 2700 J

PE = mgh
= 10x10x13.50
PE = 1350 J

Now since,

TE = PE + KE

KE = TE - PE
KE = 2700 - 1350

KE = 1350 J

Therefore, same way you can calculate KE and PE at all the other heights.

Hope this helps :)