A 5kg mass is dropped from a height above the ground after 45seconds.
I)Find the velocity of the mass as it hits the ground.
II)Calculate the kinetic energy of the mass just before it hits the ground.
III)From what height was the mass dropped.
IV)State the energy changes of the mass.
I) Using the formula:
final velocity (v) = initial velocity (u) + acceleration (a) x time (t)
Assuming the mass was dropped from rest,
u = 0m/s
a = acceleration due to gravity = 9.81 m/s^2
t = 45 seconds
v = 0 + 9.81 x 45
v = 441.45 m/s
Therefore the velocity of the mass as it hits the ground is 441.45 m/s.
II) Using the formula:
kinetic energy (KE) = 0.5 x mass x velocity^2
KE = 0.5 x 5 x (441.45)^2
KE = 0.5 x 5 x 194925.8025
Therefore the kinetic energy of the mass just before it hits the ground is 487314.50625 Joules (J).
III) Using the formula:
final height (h) = initial height (h) + initial velocity (u) x time (t) + 0.5 x acceleration (a) x time^2
Assuming the mass was dropped from rest,
u = 0m/s
a = acceleration due to gravity = 9.81 m/s^2
t = 45 seconds
final height (h) = 0 (since it hits the ground)
0 = h + 0 x 45 + 0.5 x 9.81 x (45)^2
Therefore the height from which the mass was dropped is 99,112.5 meters.
IV) The energy changes of the mass can be summarized as follows:
Potential energy (PE) at the initial height -> Kinetic energy (KE) at the final height
The potential energy is converted into kinetic energy as the mass falls due to gravity. At the final height (ground level) all the PE has been converted into KE.