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 vf = vi + gt, where vf is the final velocity, vi is the initial velocity (which is 0 for a dropped object), g is the acceleration due to gravity (which is approximately 9.81 m/s^2), and t is the time the object falls:
vf = 0 + (9.81 m/s^2)(45 s) = 441.45 m/s
Therefore, the velocity of the mass as it hits the ground is approximately 441.45 m/s.
II) Using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity:
KE = (1/2)(5 kg)(441.45 m/s)^2 = 546897.88 J
Therefore, the kinetic energy of the mass just before it hits the ground is approximately 546897.88 J.
III) Using the formula d = vit + (1/2)at^2, where d is the distance (height) the object falls, vi is the initial velocity (again, 0 for a dropped object), a is the acceleration due to gravity, and t is the time:
d = 0 + (1/2)(9.81 m/s^2)(45 s)^2 = 9918.45 m
Therefore, the mass was dropped from a height of approximately 9918.45 m.
IV) The energy changes of the mass are gravitational potential energy being converted into kinetic energy. As the mass falls, its potential energy (due to its position relative to the ground) is converted into kinetic energy (due to its motion). Just before it hits the ground, all of its potential energy has been converted to kinetic energy.