A 2kg stone is droped from the top of a 20m building
a)what potential energy does it posses
b)At what height does its potential energy become equal to its kinetic energy?
c)what is its kinetic energy just before it hits the ground?
d)With what velocity does it reach the ground?
2)what does the velocity law of conservation of energy states?
3)Define kinetic energy?
P:E 400J
PE = mgh
when half the PE is gone!
same as the original PE
set mgh = 1/2 mv^2 and solve for v
Kindly help me with the question above
ANSWER
KE=1/2 mv2
g=10m/s2
KE=O
PE =mgh
PE =2*10*20
PE=400jous
To answer these questions, we need to understand some key concepts related to potential energy, kinetic energy, and the law of conservation of energy.
1) Potential Energy:
a) The potential energy of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the object above a reference point. In this case, the mass of the stone is 2 kg and the height is 20 m. So, the potential energy is calculated as follows:
PE = 2 kg * 9.8 m/s^2 * 20 m = 392 J
b) To find the height at which the potential energy becomes equal to the kinetic energy, we know that their sum remains constant throughout the fall. At the top of the building, the stone has only potential energy, and as it falls, the potential energy gets converted to kinetic energy. When they are equal, their sum remains constant. So, when potential energy = kinetic energy, we can solve for the height using the formula PE = KE:
mgh = 1/2 * mv^2, where m is the mass, g is the acceleration due to gravity, h is the height, and v is the velocity.
Rearranging the equation, we get h = v^2 / (2g). Since the stone is dropped from rest, its initial velocity is 0. Plugging in the values:
h = (0 m/s)^2 / (2 * 9.8 m/s^2) = 0 m
c) Just before the stone hits the ground, it has only kinetic energy and no potential energy. We can calculate the kinetic energy using the formula KE = 1/2 * mv^2, where m is the mass and v is the velocity. Since we're interested in just before it hits the ground, we don't need the height. Plugging in the values:
KE = 1/2 * 2 kg * v^2 = v^2 Joules
d) To find the velocity at which it reaches the ground, we can use the equation of motion for free fall:
v^2 = u^2 + 2gh, where u is the initial velocity, g is the acceleration due to gravity, and h is the height. Since the stone is dropped from rest, the initial velocity u is 0. Substituting the values:
v^2 = 0 + 2 * 9.8 m/s^2 * 20 m
v = sqrt(2 * 9.8 m/s^2 * 20 m) = sqrt(392) ≈ 19.80 m/s
2) The Law of Conservation of Energy states that energy cannot be created nor destroyed, but it can be transformed from one form to another, or transferred between objects. In the context of velocity, it means that the total amount of mechanical energy (potential energy + kinetic energy) of a system remains constant unless external forces are acting on it.
3) Kinetic energy is the energy possessed by an object due to its motion. It is given by the equation KE = 1/2 * mv^2, where m is the mass of the object and v is the velocity. It is a scalar quantity and its unit is joules (J). The greater the mass and velocity of an object, the greater its kinetic energy.