Calculating the Work Done by Gravity on a Falling Stone

This problem asks us to find the work done by gravity on a stone with a mass of 2 kg during a specific part of its vertical fall from rest: the 5th second. Understanding the concepts of work done, gravity, and how to calculate distance in free fall is key to solving this.

Understanding Work Done and Gravity

Work done by a force is defined as the force multiplied by the distance moved in the direction of the force. In this case, the force is the gravitational force pulling the stone downwards, and the distance is how far the stone falls during the 5th second.

• The force due to gravity on an object is given by $F = mg$, where $m$ is the mass and $g$ is the acceleration due to gravity. We will use the standard value $g = 9.8 \, \text{m/s}^2$.
• Work done ($W$) is calculated as $W = F \times d$, where $F$ is the force and $d$ is the distance moved in the direction of the force. Since gravity acts downwards and the stone falls downwards, the work done by gravity will be positive.

Finding the Distance Covered in the 5th Second

The stone starts falling from rest, which means its initial velocity ($u$) is 0. To find the distance covered in the 5th second, we need to find the distance covered from $t=0$ to $t=5$ seconds and subtract the distance covered from $t=0$ to $t=4$ seconds.

Alternatively, we can use the formula for the distance covered in the n-th second of motion under constant acceleration:

$\qquad s_n = u + \frac{1}{2}a(2n - 1)$

Here:

• $s_n$ is the distance covered in the n-th second.
• $u$ is the initial velocity (0 m/s).
• $a$ is the acceleration (due to gravity, $g = 9.8 \, \text{m/s}^2$).
• $n$ is the specific second we are interested in (5th second).

Plugging in the values for the 5th second ($n=5$, $u=0$, $a=g=9.8 \, \text{m/s}^2$):

$\qquad s_5 = 0 + \frac{1}{2}(9.8)(2 \times 5 - 1)$

$\qquad s_5 = \frac{1}{2}(9.8)(10 - 1)$

$\qquad s_5 = \frac{1}{2}(9.8)(9)$

$\qquad s_5 = 4.9 \times 9$

$\qquad s_5 = 44.1 \, \text{m}$

So, the distance covered by the stone during the 5th second of its vertical fall is 44.1 meters.

Calculating the Work Done by Gravity

Now that we have the force due to gravity and the distance covered in the 5th second, we can calculate the work done by gravity.

• Mass of the stone ($m$) = 2 kg
• Acceleration due to gravity ($g$) = 9.8 m/s²
• Force due to gravity ($F$) = $mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}$
• Distance covered in the 5th second ($d$) = $s_5 = 44.1 \, \text{m}$

Work done ($W$) = Force ($F$) $\times$ Distance ($d$)

$\qquad W = 19.6 \, \text{N} \times 44.1 \, \text{m}$

Let's calculate the product:

The work done by gravity on the stone during the 5th second is 864.36 Joule.

This calculation confirms the work done by gravity for a 2 kg mass falling vertically from rest in the 5th second.

The final value for the work done by gravity is 864.36 Joule, which is one of the given options.