Question
Question: A body of mass\[5kg\]falls freely from a height of\[3m\]above the ground and penetrates\[2cm\]into t...
A body of mass5kgfalls freely from a height of3mabove the ground and penetrates2cminto the ground. Calculate the kinetic energy at a point close to the ground and the average resistive force offered by the ground.
Solution
When an object falls freely its total energy is constant and equal to sum of potential energy and kinetic energy. Using this relation we can calculate kinetic energy just above the ground. When the body hits the ground, it will exert some force on the ground and the ground will exert an equal and opposite force on it.
Formula used: E=21mv2+mgh
F=ma
Complete step by step answer:
When an object falls freely, it possesses two energies; potential energy due to its height above the ground and kinetic energy due to its motion. Therefore total energy of a freely falling body will be-E=21mv2+mgh - (1)
Here, mis the mass of the object
vis its velocity
gis acceleration due to gravity
his its height
When the object just starts falling, its potential energy is maximum and kinetic energy is 0. Just above the ground, the kinetic energy is maximum and potential energy is 0.
Therefore, we can say that,
21mv2=mgh
Kinetic energy just above the ground is equal to potential energy at height, h=3m. So, the kinetic energy will be-
K=5×10×3=150J
Kinetic energy just above the ground will be150J
According to the third law of motion, the resistive force applied by the ground on the body will be equal and opposite to the force exerted by the body on the ground. Therefore,