Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Find speed using the work-energy theorem
Reply to thread
Message
[QUOTE="Calpalned, post: 5055684, member: 525107"] [h2]Homework Statement [/h2] A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.25. What is the final speed of the crate? [h2]Homework Equations[/h2] Work energy theorem ##\Delta W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ## [h2]The Attempt at a Solution[/h2] - First I'll calculate the speed at the end of the first fifteen meters (when there is no acceleration). ##F \cdot d = (350)(15) = 5250 = ## work ##= \Delta E_k = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ## Because the initial velocity is zero, we see that ## = \frac{1}{2}mv_f^2 ## ##10500 = mv^2 ## so ##v_f = 10.458 ## m/s. - Now I'll calculate the answer (speed at 30 meters of displacement). Due to the presence of friction, kinetic and potential energies are not the only energies present. ##E_K + E_P + E_f = ## where the subscript f refers to friction ##\frac{1}{2}mv_0^2+mgh+F_N\mu kl = \frac{1}{2}mv_f^2+mgh+F_N\mu kl ## ##v_0 = 10.458## ##5250+0+0 = \frac {1}{2}mv_f^2 + 0 + 3528 ## ##v_f = 5.98## m/s [COLOR=#ff0000]The correct answer is twice what I got. What did I do wrong? [/COLOR] [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Find speed using the work-energy theorem
Back
Top