Question

A ball of mass $0.2kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2m$ while applying the force and the ball goes up to 2m in height further, find the magnitude of the force. Consider $g = 10m/{s^2}$ .
(A) 16 N
(B) 20 N
(C) 22 N
(D) 180 N

Answer 1

Potential energy is the energy possessed by a body because of the body’s position with respect to the system, the internal stresses acting on the body, the electric charge of the body and some other factors. Potential energy like any other form of energy is conserved and is closely related to the work done by the body.
Formulas used: We will be using the formula to find the potential energy of the body, $PE = mgh$ where $PE$ is the potential energy of the body, $m$ is the mass of the body, $g$ is the acceleration due to gravity of the body which is a constant and on earth is given by, $g = 10m/{s^2}$.
We will also be using the formula to find the work done on a body, $W = F \cdot s$ where $F$ is the force applied on the body, and $s$ is the displacement of the body.

Complete Step by Step answer:
We know that the potential energy of a body is the energy that a body possesses by virtue of its relative position. It is conserved with the other energy of the system, just like kinetic energy. We also know that the total work done on or by the system is equal to the change in energy of the system.
Here from the problem, we can infer that the mass of the ball is $m = 0.2kg$ and the work done by the hand causes the ball to go up a height of $h = 2m$. The hand applies a force on the ball by moving it through a distance of $s = 0.2m$.
Now by analysing the problem we can see that the ball once it is released reaches a height of $2m$ ,thus the total height of the ball would be $h' = 2 + 0.2 = 2.2m$ . By the virtue of its position with respect to the other bodies in the system, the ball possesses a potential energy, which can be given by, $PE = mgh'$
$PE = 0.2 \times 10 \times 2.2$
We also know that the ball reaches the height $h = 2m$ due to the work done by the hand which can be given by, $W = F.s$
$W = F\left( {0.2} \right)$
According to the law of conservation of energy we know that, the work done on the ball is equal to it potential energy,
$W = \Delta PE$
Substituting the known values.
$F\left( {0.2} \right) = 0.2 \times 10 \times 2.2$
Solving the equation for $F$ we get,
$F = \dfrac{{0.2 \times 10 \times 2.2}}{{0.2}}$
$\Rightarrow F = 22N$
Thus, the force applied by the hand to get the ball up in the air will be $F = 22N$ .

Hence the correct answer will be option C.

Note: The potential energy can further vary in types according to the force they are associated with, for elastic forces, the potential energy is called elastic potential energy, for the electric field by charges, the potential energy associated is called electric potential energy.