A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s.
(Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)
Thank you for the opportunity to help you with your question!
centripetal force =( m* v^2 )/r
F*r = m* v^2
kinetic energy = 1/2 * m*v ^2
kinetic energy = 1/2 *F*r
kinetic energy = 1/2 *45*1.70
kinetic energy = 38.25
after 4 seconds
kinetic energy = 153 J
Please let me know if you need any clarification. I'm always happy to answer your questions.
please go to your home page . click on promotion who has a heart icon add my name Asim Reh
or my referral link https
://www.studypool.com/Asim Reh?aid=Asim Reh
in the open window and promote me
we both will get 10$ by study pool
promote and best me dear :)
best of luck
Jul 6th, 2015
Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.