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
Power as a function of height (gravity and spring involved)
Reply to thread
Message
[QUOTE="M4573R, post: 1964962, member: 146237"] There is a "cannon" pointing downwards with a spring inside where an object is inside and compressed up 1 meter. mass of object = 50.0kg k = 49050 height = 0 at bottom of cannon. Potential energy at start = mgh + 1/2kx^2 Kinetic energy at end = 1/2mv^2 I had to first solve for the object velocity as a function of height 'y' above the bottom of the cannon. This was: v(h) = sqrt(-981y^2 - 19.6y + 1001) The question now is to solve for the amount of power being put out as a function of height. We are given the answer of: p(h) = 49y * v(h), but I am to find this on my own. I know: P = w*t, w = change in kinetic energy The change in kinetic energy can be found pretty easily, but what do I do about time? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Power as a function of height (gravity and spring involved)
Back
Top