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

This problem requires calculating the total work performed by all forces on a particle moving linearly. The particle's velocity changes with distance according to the equation \(v = \alpha \sqrt{x}\), and it moves from \(x = 0\) to \(x = d\). The objective is to determine the work done during this motion.

The work-energy principle states that the total work done on a particle is equal to its change in kinetic energy. This principle is used to find the total work done on the particle.

The kinetic energy \(K\) of the particle at any position \(x\) is defined as:

\(K = \frac{1}{2} m v^2\)

By substituting \(v = \alpha \sqrt{x}\) into the kinetic energy formula, we obtain:

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

The work done, \(W\), is calculated as the difference between the final and initial kinetic energies as the particle travels from \(x = 0\) to \(x = d\):

\(W = K_{\text{final}} - K_{\text{initial}}\)

At the initial position \(x = 0\), the velocity is \(v = \alpha \sqrt{0} = 0\), resulting in an initial kinetic energy of \(K_{\text{initial}} = 0\).

At the final position \(x = d\), the kinetic energy is \(K_{\text{final}} = \frac{1}{2} m \alpha^2 d\).

Therefore, the work done is:

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

Consequently, the total work exerted by all forces on the particle during its displacement from \(x = 0\) to \(x = d\) is:

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

This result aligns with the provided correct answer.