A particle of mass 15kg is moving with 5 m/sec along a straight line and a force of 30N is acting on it opposite to the direction of its velocity. The work done by the force during the third second of motion of the particle is:
A) $5J$
B) $0J$
C) $6J$
D) $75J$

Hint: The work done in the third second is computed by obtaining the values of work done in three seconds and work done in two seconds. The difference between the work done in the three second and the two seconds will give the work done in the third second.
Work done in the third second,
${W_{third}} = {W_3} - {W_2}$
where ${W_2}$= work done in 2 seconds and ${W_3}$= work done in 3 seconds.

Complete step by step solution:
In the problem, the given quantities are:
Mass, $m = 15kg$
Initial velocity, $u = 5m{s^{ - 1}}$
Force, $F = 30N$
The acceleration of the particle is calculated by the formula –
$F = ma$
$ \Rightarrow a = \dfrac{F}{m} = \dfrac{{30}}{{15}} = - 2m{s^{ - 2}}$
The negative sign for the acceleration is because the force is acting in the opposite direction of the motion.
Since force is acting in the direction opposite to the motion, the velocity decreases to zero in time t and there is velocity in the opposite direction.
To calculate the time t,
Final velocity, $v = 0m{s^{ - 1}}$
Initial velocity, $u = 5m{s^{ - 1}}$
Acceleration, $a = - 2m{s^{ - 2}}$
The time, t is calculated by the formula –
$v = u + at$
$0 = 5 - 2t$
$ \Rightarrow 2t = 5$
$ \Rightarrow t = \dfrac{5}{2} = 2 \cdot 5s$
Therefore, upto 2.5 seconds, the velocity of the particle reduces to zero and upto 3 seconds, the velocity is in the opposite direction due to application of the force.
Hence, the net displacement is the displacement caused in the first 2.5 seconds and displacement in the reverse direction in 3 seconds.
The formula used for calculating the displacement is:
$s = ut + \dfrac{1}{2}a{t^2}$
Displacement in 2.5 seconds,
$\Rightarrow {s_1} = 5 \times 2 \cdot 5 - \dfrac{1}{2} \times 2 \times 2 \cdot {5^2}$
$\Rightarrow {s_1} = 12 \cdot 5 - 6 \cdot 25 = 6 \cdot 25m$
The displacement in the 0.5 seconds in the opposite direction is,
$\Rightarrow {s_2} = 0 - \dfrac{1}{2} \times 2 \times 0 \cdot {5^2} = - 0 \cdot 25m$
Total displacement, $s = {s_1} + {s_2} = 6 \cdot 25 - 0 \cdot 25 = 6m$
Work done in total displacement in 3 seconds,
$\Rightarrow {W_3} = 30 \times 6 = 180J$
Using the above formula of displacement, displacement in 2 seconds,
$\Rightarrow s = 5 \times 2 - \dfrac{1}{2} \times 2 \times {2^2} = 6m$
Work done for displacement in 2 seconds,
$\Rightarrow {W_2} = 30 \times 6 = 180J$
The work done in the third second is the difference between the work done in 3 seconds and the work done in 2 seconds.
$\Rightarrow {W_{third}} = {W_3} - {W_2} = 180 - 180 = 0J$
Therefore, the work done in the third second is zero.

Hence, the correct option is Option B.

Note: The negative directions on the force and work always, indicate the relative direction of the force with respect to the motion of the body. However, the magnitude of the force and work are never negative.