Question

A body of mass\[5kg\]falls freely from a height of\[3m\]above the ground and penetrates\[2cm\]into the ground. Calculate the kinetic energy at a point close to the ground and the average resistive force offered by the ground.

Answer 1

Hint: 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: \[E=\dfrac{1}{2}m{{v}^{2}}+mgh\]
\[F=ma\]

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-\[E=\dfrac{1}{2}m{{v}^{2}}+mgh\] - (1)
Here, \[m\]is the mass of the object
\[v\]is its velocity
\[g\]is acceleration due to gravity
\[h\]is 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,
\[\dfrac{1}{2}m{{v}^{2}}=mgh\]
 Kinetic energy just above the ground is equal to potential energy at height, \[h=3m\]. So, the kinetic energy will be-
\[K=5\times 10\times 3=150J\]
Kinetic energy just above the ground will be\[150J\]
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,
\[\begin{align}
  & F=ma \\
 & \Rightarrow F=mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[a=g] \\
 & \therefore F=5\times 10=50N \\
\end{align}\]
The ground will exert a resistive force of \[50N\].

Hence, the kinetic energy of the body just above the ground is \[150J\] and the ground exerts a resistive force of \[50N\]on the body.

Note: When an object falls freely, force of gravity acts on it and the acceleration and mechanical energy of a body remains constant throughout the motion. The constant acceleration is equal to acceleration due to gravity and mechanical energy is the sum of kinetic energy and potential energy.