QJEEM24Ph3532: JEE Main 2024 Physics - Class 11 JEEM24Ph3532 Solution: A particle of mass moves on a straight line with its velocity…

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object. So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by $v = \alpha \sqrt{x}$ ,

and the kinetic energy $K$ of the particle is given by $K = \frac{1}{2} m v^2$ . We can substitute the expression for $v$ into this formula to get the kinetic energy as a function of position $x$ :

$K(x) = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x$

To find the total work done from $x = 0$ to $x = d$ , we need to compute the difference in kinetic energy between these two points:

$W = K(d) - K(0)$

At $x = d$ ,

$K(d) = \frac{1}{2} m \alpha^2 d$

At $x = 0$ , since the particle starts from this position,

$K(0) = \frac{1}{2} m \alpha^2 (0) = 0$

So, the work done $W$ is simply the kinetic energy at $x = d$ ,

$W = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d$

This matches with Option C:

$\frac{m \alpha^2 d}{2}$ .