(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

(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) Certain force acting on a 20 kg mass changes its velocity from 5 ms^-1 to 2 m s^-1, calculate the work done by the force.

(i) When an object is thrown at a certain angle, it rises to a certain height and then comes back while moving in a parabolic path. Let h be the maximum height to which it rises.
As it rises, its kinetic energy decreases but its potential energy increases.
Potential energy gained by it in rising to the highest point = work done against gravity.
= - mg x h = -mgh
As the body falls down, it gains kinetic energy = work done by the force due to gravity.
mg x h = mgh
Net work done by the force due to gravity
= - mgh + mgh = 0

(ii) Here, m = 20 kg,
u = 5 m s^-1, v = 2 m s^-1
According to work energy principle work done by a force = change in kinetic energy.
= 1/2 mv^2 - 1/2 mu^2
= 1/2 x 20 x (2)^2 - 1/2 x 20 x (5)^2
= 40 - 250
= - 210 J