an object of mass 100kg is released from a height of 4m.calculate it's kinetic energy just before it hits the ground.take g=10m/s
The KE will equal the max PE at which it started.
KE=mgh=100*g*4 joules
g is not 10m/s^2 anywhere on Earth. Fire your teacher for teaching you that. Use 9.81m/s^2
To calculate the kinetic energy of an object just before it hits the ground, we need to use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object remains constant as long as there are no external forces acting on it.
The mechanical energy of an object can be expressed as the sum of its potential energy (PE) and its kinetic energy (KE).
Potential energy (PE) is given by the formula:
PE = m * g * h
Where:
m = mass of the object (100 kg)
g = acceleration due to gravity (10 m/s^2)
h = height from which the object was released (4 m)
Kinetic energy (KE) is given by the formula:
KE = 1/2 * m * v^2
Where:
m = mass of the object (100 kg)
v = velocity of the object just before it hits the ground
To find the velocity of the object just before it hits the ground, we can use the principle of conservation of mechanical energy.
The initial mechanical energy (Ei) of the object at the topmost point is equal to the final mechanical energy (Ef) just before it hits the ground.
Ei = Ef
Since the object is released from rest at a height of 4 m, it initially has no kinetic energy, only potential energy.
Ei = PE = m * g * h
The final mechanical energy (Ef) just before it hits the ground is equal to the sum of its potential energy (PE) and kinetic energy (KE).
Ef = PE + KE
Setting Ei equal to Ef, we can solve for the kinetic energy (KE).
m * g * h = PE + KE
KE = m * g * h - PE
Substituting the given values:
KE = 100 kg * 10 m/s^2 * 4 m - 100 kg * 10 m/s^2 * 4 m
Calculating the values:
KE = 4000 kg m^2/s^2 - 4000 kg m^2/s^2
Therefore, the kinetic energy of the object just before it hits the ground is 0 J (joules).