91Ó°ÊÓ

i) Mass of ball$m=0.341\text{ k²µ}$

ii) Length of thin rod$L=0.452\text{m}$

iii) Gravitational acceleration$g=9.8{\text{m/s}}^{\text{2}}$

By using the concept of potential energy, we can find gravitational work. Gravitational work is nothing but the change in potential energy due the gravitational force.

Formula:

Gravitational potential energy is given by formula

$\begin{array}{c}U=W_g\\ =mgh\end{array}$

Workdepends on theinitial and final position. The only force that does work on the ball is the force of gravity; the force of the rod is perpendicular to the path of the ball, and so, does no work.Going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length Lof the rod, so the work done by the force of gravity is

$W_g=mgL$

$⇒W_g=0.341×9.80×0.452$

$⇒W_g=1.51\text{ J}$

In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to L, but this time the displacement is upward, opposite to the direction of the force of gravity. The work done by the force of gravity is

$W_g=−mgL$

$⇒W_g=−0.341×9.80×0.452$

$⇒W_g=−1.51\text{ J}$

Thefinal position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of gravity. The force of gravity does not work during this displacement i.e.

$W_g=0\text{ J}$

The force of gravity is conservative. The change in the gravitational potential energy of the Ball-Earth system is negative of the work done by gravity

As ball goes to final point

$\mathrm{Δ}U=−mgL$

$⇒\mathrm{Δ}U=−0.341×9.80×0.452$

$⇒\mathrm{Δ}U=−1.51\text{ J}$

For highest point

$\mathrm{Δ}U=+mgL$

$⇒\mathrm{Δ}U=+0.341×9.80×0.452$

$⇒\mathrm{Δ}U=+1.51\text{ J}$

Change in potential energy at same height is zero i.e.

$\mathrm{Δ}U=mg(L−L)$

$⇒\mathrm{Δ}U=0\text{ J}$

The change in the gravitational potential energy depends on the initial and final positions of the ball, not on its speed. The change in the potential energy is the samesince the initial and final positions are the same.