The figure below is a force displacement plot with

This preview shows page 6 - 9 out of 12 pages.

The figure below is a force-displacement plot, with the applied force F drawn above the x axis (because its work is positive) and force F s drawn below the x axis (because its work is negative).
Image of page 6

Subscribe to view the full document.

Also the net force is drawn, which is the sum of the applied constant force F and the spring force F s , F net = F - kx . The maximum speed will be achieved at the position where the net force becomes zero (equilibrium position), because up to that point the net work done on the block is positive and therefore, increases block's kinetic energy. It is so because the force F is a constant force and is NOT equal and opposite to the spring force at any moment, what is usually the case when one stretches the spring at constant (slow) speed! Here, the work done by the net force is represented by the area of a triangle, above the x axis. After that position, x o , the work done by the net force is negative, and it takes the same negative area to get to the position where kinetic energy (and speed) of the object becomes the same as the initial speed, which is in this case zero. (a) Using the figure above, the block will stop at the distance that is twice the distance where the block has maximum speed. The block has maximum speed when the net force is zero, or where k x o = F . So, the position where the block stops is: x = 2 x o = 2 F/k = 2(3.0 N)/(50 N/m) = 0.12 m. Another approach to find the position of zero speed is to use work-kinetic energy theorem. The block is at rest at the initial and final positions, so the work done by the net force is zero, because K = 0, so: k F x kx Fx W W W s F / 2 0 2 1 2 which is the same as above. (b) The work done by the applied force during displacement x is W F = F x =(3.0 N)(0.12 m) = 0.36J. (c) Because the net work is zero, the work done by the spring force is equal in magnitude and opposite in sign to the work done by the applied force, therefore W s = - W a = -0.36 J. (d) From the force-displacement plot we see that the block will have maximum speed at the position x o = x/2 = F/k which is half the distance where the block stops. Therefore, x o = x/2 = (0.12 m/2 = 0.06 m.
Image of page 7
The other approach to get position where the speed (and kinetic energy) is maximum is to express the kinetic energy, K , as a function of position, as in part (e), and take the derivative of K with respect to x , and then set the resulting expression equal to zero: x F x kx F x K kx Fx K f f / 0 2 / 2 which is the same as above. (e) Using work-kinetic energy theorem, the net work is: W net = f i f K K K K , and the final kinetic energy (at position x o ) is: . 09 . 0 ) / 50 ( 2 ) 0 . 3 ( 2 2 1 2 1 2 2 2 2 2 J m N N k F k F k F kx Fx K o o f 9. P. 7-37. (a) We first multiply the vertical axis by the mass, so that it becomes a graph of the applied force. Now, adding the triangular and rectangular ―areas‖ in the graph (for 0 x 4) gives 42 J for the work done.
Image of page 8

Subscribe to view the full document.

Image of page 9

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

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