Position 𝑥 of the particle of mass 2 kg function of time as 𝑥 = 𝑡² + 𝑡 + 1. Find out work done on the particle from 𝑡₁ = 2sec to 𝑡₂ = 3sec.

The correct option is 24 joule.

Step-by-step Explanation:

1. Understanding the given parameters:

Mass of the particle, $m=2\text{ kg}$

Position as a function of time, $x\left(t\right)=t^2+t+1$

Time interval: from $t_1=2\text{ s}$ to $t_2=3\text{ s}$

2. Finding the velocity function:

Velocity is defined as the rate of change of position with respect to time:

$v\left(t\right)=\frac{dx}{dt}$

Differentiating $x\left(t\right)=t^2+t+1$ with respect to $t$:

$v\left(t\right)=2t+1$

3. Calculating velocity at the initial and final time instants:

At initial time $t_1=2\text{ s}$:

$v_1=2\left(2\right)+1=5\text{ m/s}$

At final time $t_2=3\text{ s}$:

$v_2=2\left(3\right)+1=7\text{ m/s}$

4. Calculating the Work Done using the Work-Energy Theorem:

According to the Work-Energy Theorem, the net work done on a particle is equal to the change in its kinetic energy ($W=\Delta K$):

$W=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=\frac{1}{2}m(v_2^2-v_1^2)$

Substituting the values of $m=2\text{ kg}$, $v_1=5\text{ m/s}$, and $v_2=7\text{ m/s}$:

$W=\frac{1}{2}\times 2\times (7^2-5^2)$

$W=1\times (49-25)$

$W=24\text{ joule}$

Therefore, the work done on the particle from $t=2\text{ sec}$ to $t=3\text{ sec}$ is 24 joule.