$A$ block of mass $25 \; kg$ is raised by a $50 \; kg$ man in two different ways as shown in the figure. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of $700 \; N$,which mode should the man adopt to lift the block without the floor yielding?

(B) Mass of the block,$m = 25 \; kg$
Mass of the man,$M = 50 \; kg$
Acceleration due to gravity,$g = 10 \; m/s^2$
Force required to lift the block,$F = mg = 25 \times 10 = 250 \; N$
Weight of the man,$W = Mg = 50 \times 10 = 500 \; N$
Case $(a)$: When the man lifts the block directly,he pulls the rope upwards. By Newton's third law,the rope pulls the man downwards with a force $F = 250 \; N$. The total downward force on the floor is the sum of the man's weight and the reaction force from the rope.
Action on the floor $= W + F = 500 + 250 = 750 \; N$.
Case $(b)$: When the man lifts the block using a pulley,he pulls the rope downwards. By Newton's third law,the rope pulls the man upwards with a force $F = 250 \; N$. The net downward force on the floor is the man's weight minus the upward reaction force from the rope.
Action on the floor $= W - F = 500 - 250 = 250 \; N$.
Conclusion: Since the floor yields at $700 \; N$,the man should adopt method $(b)$ to lift the block,as the force exerted on the floor $(250 \; N)$ is less than the limit of $700 \; N$.