$(i)$ An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the total work done against the force of gravity and by the force of gravity on the object? Explain with proper mathematical expression.
$(ii)$ $A$ certain force acting on a $20 \, kg$ mass changes its velocity from $5 \, m s^{-1}$ to $2 \, m s^{-1}$. Calculate the work done by the force.

(D) $(i)$ When an object is thrown at a certain angle, it rises to a maximum height $h$ and then returns to the ground following a parabolic path. The work done against gravity while rising is $W_{against} = mgh$. The work done by gravity while falling is $W_{by} = mgh$. Since the initial and final points are on the same horizontal line, the net displacement in the vertical direction is $0$. Therefore, the total work done by the force of gravity is $W_{net} = W_{against} + W_{by} = -mgh + mgh = 0 \, J$.
$(ii)$ Given: mass $m = 20 \, kg$, initial velocity $u = 5 \, m s^{-1}$, final velocity $v = 2 \, m s^{-1}$.
According to the work-energy theorem, the work done by a force is equal to the change in kinetic energy:
$W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} \times 20 \times (2)^2 - \frac{1}{2} \times 20 \times (5)^2$
$W = 10 \times 4 - 10 \times 25$
$W = 40 - 250 = -210 \, J$.
The work done by the force is $-210 \, J$.