a body of mass 5kg falls from a height of 10m above the ground. what is the kinetic energy of the body just before it strikes the ground
answers
K.E =1/2mg'2
To determine the kinetic energy (K.E.) of the body just before it strikes the ground, you can use the formula:
K.E. = 1/2 * m * g * h
where:
m is the mass of the body (given as 5 kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the height from which the body falls (given as 10 m).
Let's calculate the kinetic energy step-by-step:
Step 1: Calculate the value of m * g * h
m * g * h = 5 kg * 9.8 m/s^2 * 10 m
= 490 kg·m^2/s^2
Step 2: Multiply the value from Step 1 by 1/2
K.E. = 1/2 * 490 kg·m^2/s^2
= 245 kg·m^2/s^2
Therefore, the kinetic energy of the body just before it strikes the ground is 245 kg·m^2/s^2.
To find the kinetic energy of the body just before it strikes the ground, you can use the formula for kinetic energy:
K.E = (1/2) * m * v^2
where K.E represents the kinetic energy, m represents the mass of the body, and v represents the velocity of the body.
In this case, the mass of the body is given as 5 kg. To find the velocity just before the body strikes the ground, you can use the principle of conservation of energy and apply the law of conservation of mechanical energy.
The potential energy of an object at a certain height is given by:
P.E = m * g * h
where P.E represents the potential energy, m represents the mass of the object, g represents the acceleration due to gravity (approximately 9.8 m/s^2), and h represents the height.
In this case, the height of the object is given as 10 m.
So, at the initial height, the potential energy is:
P.E = m * g * h
= 5 kg * 9.8 m/s^2 * 10 m
= 490 J
According to the law of conservation of mechanical energy, the potential energy at the initial height is equal to the kinetic energy just before the body hits the ground.
Therefore, the kinetic energy just before the body strikes the ground is also 490 J.