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:
In this problem, we need to calculate the total work done by all forces on a particle moving on a straight line, given its velocity changes with distance according to the equation \(v = \alpha \sqrt{x}\). The particle moves from \(x = 0\) to \(x = d\), and we aim to find the work done during this displacement.
The total work done on the particle by all forces can be found using the work-energy principle, which states that the work done on a particle equals its change in kinetic energy.
The kinetic energy \(K\) of the particle at any position \(x\) is given by:
\(K = \frac{1}{2} m v^2\)
Substituting \(v = \alpha \sqrt{x}\) into the kinetic energy formula, we get:
\(K = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x\)
The work done, \(W\), is the change in kinetic energy as the particle moves from \(x = 0\) to \(x = d\):
\(W = K_{\text{final}} - K_{\text{initial}}\)
At \(x = 0\), the velocity \(v = \alpha \sqrt{0} = 0\), so \(K_{\text{initial}} = 0\).
At \(x = d\), the kinetic energy \(K_{\text{final}} = \frac{1}{2} m \alpha^2 d\).
Hence, the work done is:
\(W = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d\)
Thus, the total work done by all the forces applied on the particle during its displacement from \(x = 0\) to \(x = d\) is:
Correct Answer:\(\frac{m \alpha^2 d}{2}\)
This matches the provided correct answer option.
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Approach Solution -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:
\( W.D = \frac{1}{2} m (\alpha \sqrt{d})^2 - \frac{1}{2} m (0)^2. \)
