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Figure Before: V 0 = 0° P < 1 of 6 After: ㄧㄢˊˋ F P Course Home - Google Chrome 1m mylabmastering.pearson.com/?courseld=12611274&key=13537117491744514152152024#/ Course Home Review Constants | Periodic Table TACTICS BOX 10.1 Calculating the work done by a constant force Direction of force relative to displacement Angles and work done Sign of W Energy transfer 0 = 0° The force is in the direction of motion. ( Figure 1) COS = 1 + The block has its greatest positive acceleration. Kinetic energy K increases the most: Maximum energy is transferred to the system. W = Fd The component of force parallel to the displacement is less than F. (Figure 2) 0 < 90° + W = Fd cos 0 The block has a smaller positive acceleration. K increases less: Moderate energy is transferred to the system. 0 = 90° There is no component of force in the direction of motion. (Figure 3) COS = 0 0 The block moves at constant speed. There is no change in K: No energy is transferred. W = 0 The component of force parallel to the displacement is opposite to the motion. ( Figure 4) 0 > 90° W = Fd cos 0 0 = 180° The force is directly opposite to the motion. ( Figure 5) cos = 1 The block slows down, and K decreases: Moderate energy is transferred out of the system. The block has its greatest deceleration. K decreases the most Maximum energy is transferred out of the system. W =-Fd A box of weight w = 2.0 N accelerates down a rough plane that is inclined at an angle = 30° above the horizontal, as shown (Figure 6). The normal force 1.7 N, the coefficient of kinetic friction between the box and the plane is = 0.30, and the displacement d of the acting on the box has a magnitude 7 = box is 1.8 m down the inclined plane. Part A What is the work Ww done on the box by the weight of the box? 66°F Sunny л 3:04 PM 3/15/2024 島

Figure Before: V 0 = 0° P < 1 of 6 After: ㄧㄢˊˋ F P Course Home - Google Chrome 1m mylabmastering.pearson.com/?courseld=12611274&key=13537117491744514152152024#/ Course Home Review Constants | Periodic Table TACTICS BOX 10.1 Calculating the work done by a constant force Direction of force relative to displacement Angles and work done Sign of W Energy transfer 0 = 0° The force is in the direction of motion. ( Figure 1) COS = 1 + The block has its greatest positive acceleration. Kinetic energy K increases the most: Maximum energy is transferred to the system. W = Fd The component of force parallel to the displacement is less than F. (Figure 2) 0 < 90° + W = Fd cos 0 The block has a smaller positive acceleration. K increases less: Moderate energy is transferred to the system. 0 = 90° There is no component of force in the direction of motion. (Figure 3) COS = 0 0 The block moves at constant speed. There is no change in K: No energy is transferred. W = 0 The component of force parallel to the displacement is opposite to the motion. ( Figure 4) 0 > 90° W = Fd cos 0 0 = 180° The force is directly opposite to the motion. ( Figure 5) cos = 1 The block slows down, and K decreases: Moderate energy is transferred out of the system. The block has its greatest deceleration. K decreases the most Maximum energy is transferred out of the system. W =-Fd A box of weight w = 2.0 N accelerates down a rough plane that is inclined at an angle = 30° above the horizontal, as shown (Figure 6). The normal force 1.7 N, the coefficient of kinetic friction between the box and the plane is = 0.30, and the displacement d of the acting on the box has a magnitude 7 = box is 1.8 m down the inclined plane. Part A What is the work Ww done on the box by the weight of the box? 66°F Sunny л 3:04 PM 3/15/2024 島

College Physics
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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Physics help!

  • what is the work Wn done on the box by the normal force?

so find Wn=____J

 

* hint given background information  "the work Ww done in the box of weight of the box is Ww= 1.8 J"

 

  • next find the work of Wfk done on the box by the force of kinetic friction? 
    give your answer in:

Wfk=___J

 

 

i have attached supporting details at the bottom.

Figure Before: V 0 = 0° P < 1 of 6 After: ㄧㄢˊˋ F
Transcribed Image Text:Figure Before: V 0 = 0° P < 1 of 6 After: ㄧㄢˊˋ F
P Course Home - Google Chrome 1m mylabmastering.pearson.com/?courseld=12611274&key=13537117491744514152152024#/ Course Home <Week 10_2: Energy Tactics Box 10.1 Calculating the Work Done by a Constant Force Learning Goal: To practice Tactics Box 10.1 Calculating the work done by a constant force. To find the work done on an object by a force, begin with a visual overview and follow Tactics Box 10.1. Recall that the work W done by a constant force F at an angle to the displacement d is WFd cos 0. The vector magnitudes F and d are always positive, so the sign of W is determined entirely by the angle between the force and the displacement. Type here to search Hi Help 1 of 6 > Review Constants | Periodic Table TACTICS BOX 10.1 Calculating the work done by a constant force Direction of force relative to displacement Angles and work done Sign of W Energy transfer 0 = 0° The force is in the direction of motion. ( Figure 1) COS = 1 + The block has its greatest positive acceleration. Kinetic energy K increases the most: Maximum energy is transferred to the system. W = Fd The component of force parallel to the displacement is less than F. (Figure 2) 0 < 90° + W = Fd cos 0 The block has a smaller positive acceleration. K increases less: Moderate energy is transferred to the system. 0 = 90° There is no component of force in the direction of motion. (Figure 3) COS = 0 0 The block moves at constant speed. There is no change in K: No energy is transferred. W = 0 The component of force parallel to the displacement is opposite to the motion. ( Figure 4) 0 > 90° W = Fd cos 0 0 = 180° The force is directly opposite to the motion. ( Figure 5) cos = 1 The block slows down, and K decreases: Moderate energy is transferred out of the system. The block has its greatest deceleration. K decreases the most Maximum energy is transferred out of the system. W =-Fd A box of weight w = 2.0 N accelerates down a rough plane that is inclined at an angle = 30° above the horizontal, as shown (Figure 6). The normal force 1.7 N, the coefficient of kinetic friction between the box and the plane is = 0.30, and the displacement d of the acting on the box has a magnitude 7 = box is 1.8 m down the inclined plane. Part A What is the work Ww done on the box by the weight of the box? 66°F Sunny л 3:04 PM 3/15/2024 島
Transcribed Image Text:P Course Home - Google Chrome 1m mylabmastering.pearson.com/?courseld=12611274&key=13537117491744514152152024#/ Course Home <Week 10_2: Energy Tactics Box 10.1 Calculating the Work Done by a Constant Force Learning Goal: To practice Tactics Box 10.1 Calculating the work done by a constant force. To find the work done on an object by a force, begin with a visual overview and follow Tactics Box 10.1. Recall that the work W done by a constant force F at an angle to the displacement d is WFd cos 0. The vector magnitudes F and d are always positive, so the sign of W is determined entirely by the angle between the force and the displacement. Type here to search Hi Help 1 of 6 > Review Constants | Periodic Table TACTICS BOX 10.1 Calculating the work done by a constant force Direction of force relative to displacement Angles and work done Sign of W Energy transfer 0 = 0° The force is in the direction of motion. ( Figure 1) COS = 1 + The block has its greatest positive acceleration. Kinetic energy K increases the most: Maximum energy is transferred to the system. W = Fd The component of force parallel to the displacement is less than F. (Figure 2) 0 < 90° + W = Fd cos 0 The block has a smaller positive acceleration. K increases less: Moderate energy is transferred to the system. 0 = 90° There is no component of force in the direction of motion. (Figure 3) COS = 0 0 The block moves at constant speed. There is no change in K: No energy is transferred. W = 0 The component of force parallel to the displacement is opposite to the motion. ( Figure 4) 0 > 90° W = Fd cos 0 0 = 180° The force is directly opposite to the motion. ( Figure 5) cos = 1 The block slows down, and K decreases: Moderate energy is transferred out of the system. The block has its greatest deceleration. K decreases the most Maximum energy is transferred out of the system. W =-Fd A box of weight w = 2.0 N accelerates down a rough plane that is inclined at an angle = 30° above the horizontal, as shown (Figure 6). The normal force 1.7 N, the coefficient of kinetic friction between the box and the plane is = 0.30, and the displacement d of the acting on the box has a magnitude 7 = box is 1.8 m down the inclined plane. Part A What is the work Ww done on the box by the weight of the box? 66°F Sunny л 3:04 PM 3/15/2024 島
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