Question

A ball of mass $2kg$ is dropped from a height. What is the work done by its weight in $2$ seconds after the ball is dropped?

Answer 1

Hint : In this problem equation of motion that is, $S = ut + \dfrac{1}{2}a{t^2}$ will be used to find the height, then work done by the ball of mass $2kg$ in time $2$ seconds can be calculated by using the formula $W = mgh$ .S.I unit of work is joules $\left( J \right)$ .It should be noted that initial velocity is zero for freely falling body.

Formulas used:
$S = ut + \dfrac{1}{2}a{t^2}$ ………… $\left( 1 \right)$
Work is said to be done when force applied on a body makes the body to move,
Therefore , $W = FS$ ………….$\left( 2 \right)$
According to the Newton’s second law
$F = ma$ …………. $\left( 3 \right)$
Substituting the equation $\left( 3 \right)$ in the equation $\left( 2 \right)$
$W = mgh$

Complete step-by-step solution:
Given:
Time, $t = 2s$ .
Mass, $m = 2kg$ .
For freely falling body,
Initial velocity $u = 0$ ($\because$ because kinetic energy is zero at maximum height).
Therefore the kinetic energy will be zero at the maximum height and the potential energy will be maximum.
$S = h$ And $a = g$ ($g =$ acceleration due to gravity).
We know that, from the equation of motion the displacement of the ball is given by,
$S = ut + \dfrac{1}{2}a{t^2}$ .
Substituting the given data in above equation, we get
$h = ut + \dfrac{1}{2}g{t^2}$
$h = 0 \times 2 + \dfrac{1}{2} \times 9.8 \times {2^2}$
$h = \dfrac{1}{2} \times 9.8 \times 4$
On simplifying the above equation, we get
$h = 19.6m$
We know that, work done is given by
$W = mgh$ . ($\because$ the work done will be equal to the potential energy)
On substituting, the above equation becomes
$W = 2 \times 9.8 \times 19.6$
On calculating, we get work done as
$W = 348.16J$
Hence, the work done by the ball of mass $2kg$ to reach a height of $19.6m$ in time $2s$ is $348.16J$

Note: For a freely falling body the initial velocity should be considered as zero because at maximum height the potential energy will be maximum and kinetic energy will be zero. As the object falls the potential energy will be converted to kinetic energy. The unit for the work done can be joules $\left( J \right)$ or electron volt $\left( {eV} \right)$ ,the relation between electron volt and joules is $1eV = 1.6 \times {10^{ - 19}}J$ .