An object is falling freely from a height x. After it has fallen to a height x/2 , it will possess.
half potential and half kinetic energy
Understanding Energy Transformation During Free Fall
When an object is falling freely from a height, its energy transforms from potential energy to kinetic energy. This transformation happens while the total mechanical energy remains constant, assuming negligible air resistance. This concept is central to understanding the Energy in Free Fall.
Initial State and Energy
Let the object of mass 'm' start falling freely from a height 'x' above the ground. Since it's falling freely, its initial velocity is zero.
- Initial height \(h_i = x\)
- Initial velocity \(v_i = 0\)
The initial potential energy (\(PE_i\)) is the energy it possesses due to its position in the gravitational field. This is also known as gravitational potential energy.
\[PE_i = mgh_i = mgx\]The initial kinetic energy (\(KE_i\)) is the energy it possesses due to its motion. Since the initial velocity is zero, the initial kinetic energy is also zero.
\[KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}m(0)^2 = 0\]The total initial mechanical energy (\(E_i\)) is the sum of initial potential and kinetic energy.
\[E_i = PE_i + KE_i = mgx + 0 = mgx\]Energy at Half the Initial Height (x/2)
Now, consider the object when it has fallen to a height of \(x/2\) above the ground. At this point, the object has gained some velocity due to falling.
- Final height \(h_f = x/2\)
- Let the velocity at this height be \(v_f\)
The potential energy (\(PE_f\)) at this height is:
\[PE_f = mgh_f = mg\left(\frac{x}{2}\right)\]The kinetic energy (\(KE_f\)) at this height is:
\[KE_f = \frac{1}{2}mv_f^2\]To find \(KE_f\), we need to determine \(v_f^2\). We can use the equations of motion for an object under constant acceleration due to gravity. The object has fallen a distance of \(x - x/2 = x/2\). Using the equation \(v^2 = u^2 + 2as\), where \(u=0\), \(a=g\), and \(s=x/2\):
\[v_f^2 = 0^2 + 2g\left(\frac{x}{2}\right) = gx\]Now substitute this value of \(v_f^2\) into the kinetic energy formula:
\[KE_f = \frac{1}{2}m(gx) = \frac{1}{2}mgx\]So, at height \(x/2\), the potential energy is \(mgx/2\) and the kinetic energy is \(mgx/2\).
Comparing Potential and Kinetic Energy at x/2
Let's compare the potential energy and kinetic energy at height \(x/2\):
- Potential Energy \(PE_f = \frac{1}{2}mgx\)
- Kinetic Energy \(KE_f = \frac{1}{2}mgx\)
As you can see, at height \(x/2\), the potential energy is equal to the kinetic energy.
\[PE_f = KE_f\]The total mechanical energy at height \(x/2\) is the sum of \(PE_f\) and \(KE_f\):
\[E_f = PE_f + KE_f = \frac{1}{2}mgx + \frac{1}{2}mgx = mgx\]This confirms the principle of conservation of energy, as the total energy \(E_f\) is equal to the initial total energy \(E_i\). This conservation of energy is a fundamental aspect of Energy in Free Fall.
Conclusion on Energy Distribution at x/2
At the height \(x/2\), the object possesses potential energy equal to \(mgx/2\) and kinetic energy equal to \(mgx/2\). The total energy is \(mgx\). This means that at this point, the potential energy is half of the total initial energy (\(mgx\)), and the kinetic energy is also half of the total initial energy (\(mgx\)). The transformation of gravitational potential energy into kinetic energy during the free fall is exactly such that at height \(x/2\), the energy is split equally.
Therefore, when the object has fallen to a height \(x/2\), it will possess half potential energy and half kinetic energy. This demonstrates the dynamic balance between potential energy and kinetic energy during Energy in Free Fall.
Important Questions from Gravitation
Which one of the following statement is true for the relation, \(F= \frac{{G{m_1}{m_2}}}{{{r^2}}}\) ?
(All symbols have their usual meanings)The free-fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is
A planet has a mass M 1and radius R 1. The value of acceleration due to gravity on its surface is g 1. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?
Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be
Suppose the force of gravitation between two bodies of equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become
