Question

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)

Hint:

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.

Answer 1

Step 1: Understanding Work Done:
Work done (\(W\)) by a force depends on the force applied and the displacement of the object along the direction of the force. - If the force vector \(\vec{F}\) is constant and the displacement is \(\vec{d}\), then work done is \(W = \vec{F} \cdot \vec{d} = Fd\cos\phi\), where \(\phi\) is the angle between the force and displacement vectors. - If the force is applied along a path, the work done is the integral of the force component along the path, \(W = \int \vec{F} \cdot d\vec{s}\).
Step 2: Calculating Work Done for Each Case:
The net displacement from A to B is a straight line of length \(d = 20\) m.
Case 1: The force F is applied tangentially along the semi-circular path AJB. The length of the path is the circumference of the semi-circle, \(s = \pi r\). The radius is \(r = AB/2 = 20/2 = 10\) m. So, path length \(s = \pi \times 10 = 10\pi\) m. Since the force is always in the direction of motion, the work done is force \(\times\) distance. \[ W_1 = F \times s = F \times 10\pi \approx 31.4 F \] Case 2: The constant force F is applied horizontally, parallel to the displacement AB. The displacement is \(d = 20\) m. The angle between force and displacement is \(\phi = 0^\circ\). \[ W_2 = F \times d \times \cos(0^\circ) = F \times 20 \times 1 = 20 F \] Case 3: The constant force F is applied at an angle \(\theta\) to the horizontal displacement AB. The displacement is \(d = 20\) m. The angle between force and displacement is \(\theta\). \[ W_3 = F \times d \times \cos\theta = F \times 20 \times \cos\theta \] Given that \(0^\circ<\theta<90^\circ\), we know that \(0<\cos\theta<1\). Therefore, \(W_3\) is less than \(20F\). For example, if \(\theta = 60^\circ\), \(W_3 = 20F \times 0.5 = 10F\).
Step 3: Comparing the Work Done:
We have: - \(W_1 \approx 31.4 F\) - \(W_2 = 20 F\) - \(W_3 = (20 \cos\theta) F\), which is less than \(20F\). Comparing the three values, we get: \(W_3<W_2<W_1\).
The ascending order is Case 3, Case 2, Case 1.