A particle of mass 0.01 kg travels along a curve with velocity given by 4î + 16k̂ ms⁻¹ . After some time, its velocity becomes 8î + 20ĵ ms⁻¹ due to the action of a conservative force. The work done on particle during this interval of time is

The correct option is 0.96 J.

Step-by-Step Explanation:

According to the Work-Energy Theorem, the work done by all the forces acting on a particle is equal to the change in its kinetic energy:

$W=\mathrm{\Delta K}=K_{\text{final}}-K_{\text{initial}}$

The kinetic energy of a particle of mass $m$ moving with velocity $v$ is given by:

$K=\frac{1}{2}m{v}^2$

where $v^2=v·v$ is the square of the magnitude of the velocity vector.

Given details:
Mass of the particle, $m=0.01\text{kg}$
Initial velocity, $v_{\text{i}}=4\hat{i}+16\hat{k}{\text{m s}}^{-1}$
Final velocity, $v_{\text{f}}=8\hat{i}+20\hat{j}{\text{m s}}^{-1}$

Step 1: Calculate the square of the initial velocity magnitude ($v_{\text{i}}^2$)

$v_{\text{i}}^2=4^2+{16}^2$

$v_{\text{i}}^2=16+256=272{\text{m}}^2{\text{s}}^{-2}$

Step 2: Calculate the square of the final velocity magnitude ($v_{\text{f}}^2$)

$v_{\text{f}}^2=8^2+{20}^2$

$v_{\text{f}}^2=64+400=464{\text{m}}^2{\text{s}}^{-2}$

Step 3: Compute the work done on the particle

$W=\frac{1}{2}m(v_{\text{f}}^2-v_{\text{i}}^2)$

Substitute the given values into the equation:

$W=\frac{1}{2}\times 0.01\times (464-272)$

$W=0.005\times 192$

$W=0.96\text{J}$

Therefore, the work done on the particle during this time interval is 0.96 J.