Please, Help with question #2Thank you. **Question 2:**Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping. How much work did the workman (through the use of the rope) do to the sled?- **A)** \(50,000 \sin(\theta) \, [J]\)- **B)** \(50,000 \cos(\theta) \, [J]\) *(Correct Answer)*- **C)** \(50,000 \, [J]\)- **D)** \(200 \, [J]\)- **E)** None of the above*Explanation:*The problem asks for the calculation of work done by the workman on the sled, using the force applied and the distance moved. The key to the answer depends on the angle \(\theta\) at which the force is applied, making option B correct, as it involves calculating the component of the force in the direction of the movement: \(W = F \cdot d \cdot \cos(\theta)\). In this question, option B uses the cosine of the angle, aligning with the direction of movement. ### Problem and Solution Explanation#### Problem Description:The question addresses the dynamics of a sled being pulled across an asphalt road at an angle using a rope. The key points are:- A workman exerts a force of 1000 N at an angle \( \theta \) above the horizontal.- The sled has a mass \( m \).- Coefficients of friction are given as \( \mu_s = 1.00 \) (static) and \( \mu_k = 0.70 \) (kinetic).The problem asks which expression calculates the maximum static friction \( f_s^{max} \) for the scenario described, given the options:A) \( mg \)B) \( mg - 1000\cos(\theta) \)C) \( mg + 1000\cos(\theta) \)D) \( mg - 1000\sin(\theta) \)E) \( mg + 1000\sin(\theta) \)The chosen answer is D.#### Solution Breakdown:1. **Diagram Explanation**: - The force \( T \) (1000 N) is applied at an angle \( \theta \). - The vertical forces are \( N \) (normal force) and \( mg \) (gravitational force).2. **Equations Used**: From equilibrium of vertical forces: \[ N + 1000\sin(\theta) = mg \] Solving for \( N \): \[ N = mg - 1000\sin(\theta) \]3. **Calculating Maximum Static Friction (\( f_s^{max} \))**: Using the formula for static friction: \[ f_s^{max} = \mu_s N = 1.00 \times (mg - 1000\sin(\theta)) \] Simplifying: \[ f_s^{max} = mg - 1000\sin(\theta) \]4. **Conclusion**: The expression \( mg - 1000\sin(\theta) \) correctly represents the maximum static friction, making option D the correct answer.This detailed explanation helps in understanding the application of forces, friction, and equilibrium in solving physics problems related to motion and static equilibrium.

Answered: Image /qna-images/answer/c9e7d63b-f96d-4007-85c3-a048eb7875ea.jpg

2) Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping. How much work did the workman (through the use of the rope) do to the sled? A) 50,000sin(0) [J] XB) 50,000cos(0) [J] C) 50,000 [J] 200 [J] E) None of the above

2) Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping. How much work did the workman (through the use of the rope) do to the sled? A) 50,000sin(0) [J] XB) 50,000cos(0) [J] C) 50,000 [J] 200 [J] E) None of the above

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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