Question

An object of mass $40{\rm{ kg}}$ is raised to a height of 5m above the ground. What is the potential to fall, find its kinetic energy when it is half way done?

Answer 1

From the concept of energy conservation of the given object, we can say that the summation of kinetic and potential energies of the object at the maximum height is equal to the summation of its kinetic and potential energies when it is half way done.

Complete step by step answer:
Given:
The mass of the object is $m = 40{\rm{ kg}}$.
The height from ground to which the object is raised is $h = 5{\rm{ m}}$.
We have found the potential energy of the body when it is at distance h from the ground. Also, we have to find the kinetic energy of the object when it is half way done.
Let us write the expression for potential energy of the given object when it is at a distance h from the ground.
$P = mgh$
Here g is acceleration due to gravity and value is given by $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$.
On substituting $40{\rm{ kg}}$ for m, $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ for g, and $5{\rm{ m}}$ for h in the expression, we get:

$$P = \left( {40{\rm{ kg}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)\left( {5{\rm{ m}}} \right)\\\ = 1962{\rm{ J}}$$

Let us consider the height h as the initial state of the object and height $\dfrac{h}{2}$ as final state.
From the concept of law of energy conservation, we can write:
${K_i} + {P_i} = {K_j} + {P_j}$……(1)
Here ${K_i}$ is the initial kinetic energy, ${K_j}$ is the final kinetic energy, ${P_i}$ is the initial potential energy and ${P_j}$ is the final kinetic energy of the object.
We can write the expression for final potential energy of the object as below:
${P_j} = mg\dfrac{h}{2}$
On substituting $40{\rm{ kg}}$ for m, $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ for g, and $5{\rm{ m}}$ for h in the expression, we get:

$${P_j} = \left( {40{\rm{ kg}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)\dfrac{{5{\rm{ m}}}}{2}\\\ = 981{\rm{ J}}$$

We know that the initial object is not moving so its initial kinetic energy is zero.
Substitute 0 for ${K_i}$, $1962{\rm{ J}}$ for ${P_i}$ and $981{\rm{ J}}$ for ${P_j}$ in equation (1).

$$0 + 1962{\rm{ }}{{\rm{J}}_i} = {K_j} + 981{\rm{ J}}\\\ {K_j} = 981{\rm{ J}}$$

Therefore, the potential to fall is $1962{\rm{ J}}$ and kinetic energy when the object is half way done is $981{\rm{ J}}$.

Note: Kinetic energy of the object is energy associated with it due to its motion, whereas potential energy of the object is energy stored in the object when it is lifted against the action of gravity.