{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MASTERING PHYSICS CHAPTER 4

MASTERING PHYSICS CHAPTER 4 - 4.1 Solve A force is...

Info icon This preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a point of contact between the object and its environment. (iii) A very few forces, such as gravity and magnetism, are long-range forces that require no contact. (iv) A force is a vector quantity , having both a magnitude (or size) and a direction. (v) When multiple forces act on an object, the forces combine through vector addition to give a net force r r F F i net = .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.2. Visualize: Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls.
Image of page 2
4.3. Visualize:
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.4. Model: Assume friction is negligible compared to other forces. Visualize:
Image of page 4
4.5. Visualize: Please refer to Figure Ex4.5. Solve: Mass is defined to be m = 1 slope of the acceleration-versus-force graph A larger slope implies a smaller mass. We know m 2 = 0.20 kg, and we can find the other masses relative to m 2 by comparing their slopes. Thus m m m m 1 2 1 2 1 1 1 5 2 2 5 0 40 0 40 0 40 0 20 0 08 = = = = = = = × = / / . . . . . slope 1 slope 2 slope 2 slope 1 kg kg Similarly, m m m m 3 2 3 2 1 1 1 2 5 5 2 2 50 2 50 2 50 0 20 0 50 = = = = = = = × = / / . . . . . slope 3 slope 2 slope 2 slope 3 kg kg Assess: From the initial analysis of the slopes we had expected m 3 > m 2 and m 1 < m 2 . This is consistent with our numerical answers.
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.6. Model: An object’s acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F , two rubber bands produce a force 2 F , and so on. Because F a and two rubber bands (force 2 F ) produce an acceleration of 1.2 m/s 2 , four rubber bands will produce an acceleration of 2.4 m/s 2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2 m ). Using F ma = , 2 F = (2 m ) a a = F / m = 0.6 m/s 2
Image of page 6
4.7. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes acceleration . That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force r F net on an object of mass m causes the object to undergo an acceleration: r r a F m = net The acceleration vector and the net force vector must point in the same direction.
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.8. Visualize: Solve: Newton’s second law is F ma = . When F = 2 N, we have 2 N = (0.5 kg) a , hence a = 4 m/s 2 . After repeating this procedure at various points, the above graph is obtained.
Image of page 8
4.9. Visualize: Please refer to Figure Ex4.9. Solve: Newton’s second law is F ma = . We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2 . Newton’s second law yields m = 0.25 kg.
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.10.
Image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern