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

The Correct Option is C

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: \( \frac{m \alpha^2 d}{2}. \)