Question

A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha \sqrt{x}$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$, will be:

• A. $\dfrac{m}{2\alpha^2 d}$
• B. $\dfrac{md}{2\alpha^2}$
• C. $\dfrac{m\alpha^2 d}{2}$
• D. $2m\alpha^2 d$

Answer 1

Answer: (C)

$\dfrac{m\alpha^2 d}{2}$

Answer 2

Step 1: Velocity equation The velocity of the particle is given as:

$v = \alpha \sqrt{x}.$

At $x = 0$, the velocity is:

$v = 0.$

At $x = d$, the velocity becomes:

$v = \alpha \sqrt{d}.$

Step 2: Work-Energy Theorem The work done by all forces is equal to the change in kinetic energy:

$W.D = K_f - K_i,$

where:

$K = \frac{1}{2} mv^2.$

Substitute the velocities:

$W.D = \frac{1}{2} m (\alpha \sqrt{d})^2 - \frac{1}{2} m (0)^2.$

Simplify:

$W.D = \frac{1}{2} m (\alpha^2 d) - 0.$

$W.D = \frac{m \alpha^2 d}{2}.$

Final Answer: $\frac{m \alpha^2 d}{2}.$