cant copy graph Two students, Alice and Bob, decide to compute the power that the Earth's gravitational field expends on a block of mass \(m\) as the block slides down a frictionless inclined plane. Alice reasons: "The gravitational force pulling the block down the incline is \(F=m g \sin \theta\). The block's velocity at any given height \(h\) from the top of the incline is \(v=\sqrt{2 g h}\). Power is defined as force \(\times\) velocity. Therefore, the power is \(P=m g \sin \theta \sqrt{2 g h}\)." Bob reasons: "Power is \(\Delta W / \Delta t\). By the work-energy theorem, the change in work is the change in kinetic energy, but without friction \(\Delta W=m g h\). The change in time is \(\Delta t=\Delta v / a, \Delta v=\sqrt{2 g h}\) and \(a=g \sin \theta\). So \(\Delta t=\sqrt{2 g h} /(g \sin \theta)\). Therefore, \(P=m g h \times \frac{g \sin \theta}{\sqrt{2 g h}}=m g \sin \theta \sqrt{\frac{g h}{2}}\)." Alice and Bob look at each other and scratch their heads. Who is correct? \(\begin{array}{llll}\text { i. Alice } & \text { ii. Bob } & \text { iii. Neither iv. Both } & \text { v. The problem is imprecisely worded. }\end{array}\) (A) \(\mathrm{i}\) (B) ii (C) iii (D) iv (E) iv and \(v\)

Step by step solution

01

Review Alice's reasoning

Alice used the following equations: 1. Gravitational force along the incline: \(F = mg\sin\theta\) 2. Velocity at height \(h\): \(v = \sqrt{2gh}\) 3. Power: \(P = F \times v\) By substituting equations 1 and 2 into equation 3, Alice got \(P = m g \sin\theta \sqrt{2gh}\).

02

Review Bob's reasoning

Bob used the following relations: 1. Power: \(P = \Delta W / \Delta t\) 2. Without friction, change in work \(\Delta W = mgh\) 3. Change in time: \(\Delta t = \Delta v / a\), with \(\Delta v = \sqrt{2gh}\) and \(a = g\sin\theta\) By substituting equations 2 and 3 into equation 1, Bob got \(P = mg\sin\theta \sqrt{\frac{gh}{2}}\).

03

Compare their solutions

Alice's solution: \(P = mg\sin\theta\sqrt{2gh}\) Bob's solution: \(P = mg\sin\theta\sqrt{\frac{gh}{2}}\) Both equations have the same structure and same variables but different factors in the square root.

04

Identify the incorrect step in Bob's reasoning

Bob's mistake lies in the step where he calculates the change in time \(\Delta t\). He uses the equations \(\Delta t = \Delta v/a\) with \(\Delta v = \sqrt{2gh}\) and \(a = g\sin\theta\). However, this is incorrect because one should use \(v = \frac{d}{\Delta t}\), where \(d\) is the distance traveled and \(v\), \(a\) are evaluated along the incline, instead of \(\Delta t = \Delta v/a\).

05

Determine the correct answer

Since Alice's derivation is correct and Bob's is flawed, the answer should be: (A) i. Alice is correct.

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