You are a fitness trainer developing an exercise that involves pulling a heavy crate across a rough surface. You attach a rope to a crate of mass ?=45.0 kg and place it on a carpeted track. Using your physics expertise, you have already determined that the coefficient of kinetic friction between the crate and the carpet is ?k=0.470. To test the apparatus, you have one of your employees pull on the rope as hard as she can. If she pulls with a force of ?=315 N along the direction of the rope, while the rope makes an angle of ?=23.0° with the floor,what is the acceleration ? of the box? Use ?=9.80 m/s2. The image depicts a wooden crate placed on a horizontal surface. A force \( F \) is applied to the crate through a rope attached to it. The direction of the force \( F \) forms an angle \( \theta \) with the horizontal axis. Key elements in the diagram include:1. **Crate (M):** The crate is shown as a wooden box, presumed to have a mass \( M \).2. **Force (F):** A force is applied at an angle \( \theta \) to the horizontal line, indicating the direction and application of the force.3. **Angle (\( \theta \)):** The angle formed between the horizontal direction and the applied force \( F \).4. **Surface (\( \mu_k \)):** The coefficient of kinetic friction (\( \mu_k \)) is represented on the surface, suggesting that friction is acting on the moving crate.This diagram could be used to explain concepts related to forces, friction, and motion in physics.

The force equation in the horizontal direction is given by, ΣFx=Fx-fk (1) Here, ΣFx is the net force in horizontal direction, Fx is the horizontal component of the applied force and fk is the force of friction. The net force is also given by, ΣFx=Ma (2) Here, a is the acceleration of the box. The expression for Fx is given by, Fx=Fcosθ (3) The force of friction is given by, fk=μkN Here, N is the normal force. The normal force is given by, N=Mg-Fy=Mg-Fsinθ Here, Fy is the vertical component of the applied force. Thus, the force of friction is, fk=μkMg-Fsinθ (4)Putting equations (2) to (4) in equation (1), Ma=Fcosθ-μkMg-Fsinθa=Fcosθ-μkMg-FsinθM=315 Ncos23.0°-0.47045.0 kg9.80 m/s2-315 Nsin23.0°45.0 kg=3.12 m/s2

Answered: You are a fitness trainer developing an…

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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You are a fitness trainer developing an exercise that involves pulling a heavy crate across a rough surface. You attach a rope to a crate of mass ?=45.0 kg and place it on a carpeted track. Using your physics expertise, you have already determined that the coefficient of kinetic friction between the crate and the carpet is ?k=0.470.

To test the apparatus, you have one of your employees pull on the rope as hard as she can. If she pulls with a force of ?=315 N along the direction of the rope, while the rope makes an angle of ?=23.0° with the floor,what is the acceleration ? of the box? Use ?=9.80 m/s2.

The image depicts a wooden crate placed on a horizontal surface. A force \( F \) is applied to the crate through a rope attached to it. The direction of the force \( F \) forms an angle \( \theta \) with the horizontal axis.

Key elements in the diagram include:

1. **Crate (M):** The crate is shown as a wooden box, presumed to have a mass \( M \).

2. **Force (F):** A force is applied at an angle \( \theta \) to the horizontal line, indicating the direction and application of the force.

3. **Angle (\( \theta \)):** The angle formed between the horizontal direction and the applied force \( F \).

4. **Surface (\( \mu_k \)):** The coefficient of kinetic friction (\( \mu_k \)) is represented on the surface, suggesting that friction is acting on the moving crate.

This diagram could be used to explain concepts related to forces, friction, and motion in physics.

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