A stone is tied to a string and displaced from A to B by application of a constant force F in three different ways as shown in the diagram below. Arrange the three cases in ascending order of the work done by the force. (Given AJB is a semi-circle, 0\(^\circ\)<\(\theta\)<90\(^\circ\) and AB = 20 m)
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Work done is maximized when the force is applied exactly along the path of motion. Any angle between the force and the direction of motion reduces the effective work done for a given displacement.
Step 1: Understanding the Concept:
Work done is defined as the product of the component of force in the direction of displacement and the magnitude of the displacement: \(W = F \cdot s \cdot \cos(\theta)\). Step 2: Key Formula or Approach:
\(W = \text{Force} \times \text{Displacement in the direction of force}\). Step 3: Detailed Explanation:
Let \(F\) be the constant force magnitude.
1. Case 2: Displacement is along the straight line AB. The force \(F\) is in the same direction as displacement.
\(s = 20 \text{ m}\). \(W_{2} = F \times 20 = 20F\).
2. Case 3: Displacement is along AB (20 m), but the force acts at an angle \(\theta\).
\(W_{3} = F \cdot \cos(\theta) \times 20 = 20F \cos(\theta)\). Since \(0<\theta<90^{\circ}\), \(\cos(\theta)<1\). Thus \(W_{3}<W_{2}\).
3. Case 1: The stone moves along the semi-circle arc AJB. The force \(F\) is shown acting along the tangent at every point (implied by the arrow following the curve).
Distance along arc = \(\pi \times \text{radius} = \pi \times \frac{20}{2} = 10\pi \approx 31.4 \text{ m}\).
\(W_{1} = F \times 10\pi \approx 31.4F\).
Comparing values: \(20F \cos(\theta)<20F<31.4F\). Step 4: Final Answer:
The ascending order is Case 3 < Case 2 < Case 1.
