Correct Option: (c)

Let’s analyze the forces in both cases. The man applies a force $F$ at an angle $\theta$ with the horizontal.

1. Comparing Work Done (Statement II)

The work done by the man is defined as the product of the force component in the direction of displacement and the displacement $d$. In both pushing and pulling, the horizontal component of the force is $F \cos\theta$.

$$ W_{man} = (F \cos\theta) \cdot d $$

Since $F$, $\theta$, and $d$ are identical in both cases, both men do an equal amount of work.

2. Comparing Friction and Power (Statement V)

Normal Force ($N$):

• Pushing: The vertical component of $F$ points down. $N_{push} = mg + F\sin\theta$.
• Pulling: The vertical component of $F$ points up. $N_{pull} = mg – F\sin\theta$.

Since friction $f = \mu N$, and $N_{push} > N_{pull}$, the friction during pushing is greater: $f_{push} > f_{pull}$.

Net Acceleration ($a$):

$$ a = \frac{F\cos\theta – f}{m} $$

Because the retarding friction force is smaller in pulling, the net acceleration is higher: $a_{pull} > a_{push}$.

Time ($t$) and Power ($P$):

To cover the same distance $d$ starting from rest, time taken is $t = \sqrt{\frac{2d}{a}}$. Since acceleration is higher for pulling, time taken is less: $t_{pull} < t_{push}$.

Average power delivered is $P = \frac{W_{man}}{t}$. Since work is equal but time is smaller for the second man (puller), he delivers more power.

Conclusion: Statement II (Equal Work) and Statement V (Second man delivers more power) are correct.