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Q6.5(d):

Answer the following : (d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?

Solution :

Step 1: Theoretical Foundation of Work

In physics, work $W$ done by a constant force $F$ is defined as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$. Mathematically, this is expressed as:

$W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos(\theta)$

where $\theta$ is the angle between the direction of the force and the direction of displacement. [Per the definition of mechanical work].

Step 2: Analysis of Case (i) - Carrying a Mass

In Case (i), the man carries a mass $m = 15\text{ kg}$ while walking a horizontal distance $d = 2\text{ m}$.

  • Force applied: To hold the mass, the man must exert an upward vertical force $F_v = mg$ to counteract gravity. $F_v = 15\text{ kg} \times 9.8\text{ m/s}^2 = 147\text{ N}$.
  • Displacement: The displacement vector $\vec{d}$ is horizontal.
  • Angle ($\theta$): The angle between the upward force $\vec{F_v}$ and the horizontal displacement $\vec{d}$ is $90^\circ$.
  • Calculation: $W_i = F_v \cdot d \cdot \cos(90^\circ) = 147 \cdot 2 \cdot 0 = 0\text{ J}$.

[Justification: The work done by the man against gravity is zero because the force is perpendicular to the displacement.]

Step 3: Analysis of Case (ii) - Pulling a Mass

In Case (ii), the man pulls a rope horizontally to lift a mass $m = 15\text{ kg}$ hanging over a pulley.

  • Force applied: To lift the mass, the man must apply a tension $T$ in the rope equal to the weight of the mass. $T = mg = 147\text{ N}$.
  • Displacement: When the man walks a distance $d = 2\text{ m}$, the hanging mass is lifted vertically by a height $h = 2\text{ m}$.
  • Angle ($\theta$): The force applied by the man on the rope is in the direction of the rope, and the displacement of the mass is in the direction of the tension (upwards). The angle $\theta$ between the tension force and the displacement of the mass is $0^\circ$.
  • Calculation: $W_{ii} = T \cdot h \cdot \cos(0^\circ) = 147 \cdot 2 \cdot 1 = 294\text{ J}$.

Step 4: Comparative Visualization

m=15kg Case (i): W = 0 J Case (ii): W = 294 J

Step 5: Conclusion

Comparing the two results:

Case Work Done
Case (i) $0\text{ J}$
Case (ii) $294\text{ J}$

Final Solution: The work done is greater in Case (ii), where the man performs $294\text{ J}$ of work, whereas in Case (i), the work done against gravity is $0\text{ J}$.


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