Home » Blog » Question of The Day » Problem no.10, Topic: Work Power and Energy
Tags: Physics, problems, question, Question of The Day, Work power and Energy
neema December 20, 2010 at 1:21 pm
my ans is mgsin(theta)-kR(theta)^2-2mgcos(theta) plz let me know if it is correct….
neema December 20, 2010 at 1:35 pm
my first ans is wrong…the ans should be (kR(theta)^2)+(mgsin(theta))
neema December 20, 2010 at 1:39 pm
my ans is kR(theta)^2+(mgsin(theta)) plz let me know if it is correct….
Anudeep December 21, 2010 at 9:34 pm
isnt the answer so given wrong, how can the tangential force F applied be considered in the contact force
rishabh March 2, 2011 at 12:45 am
sir why don’t you give the SOLUTION..
abhishek.hiremath June 22, 2011 at 3:26 pm
wud somebody mind giving d answers wid d method of solving..:!!
Dipak Savaliya September 11, 2011 at 12:38 pm
it’s answer May be 3mgsin(theta)+kR(theta)^2+2F(theta)
lakshya September 11, 2011 at 8:39 pm
Umm…i dont mind giving the solution… The first equation 1) mgsin theta -N = mv^2/R 2)The tangential forces on the body will give the equation for its acceleration*mass. At an angle theta from the centre :- F – (kx+mgcos theta) = m.a Here x = R.theta…from angle = arc/radius Then…instead of a…we put v.(dv/dx) Now it becomes integral of [F – kRtheta – mgcos theta].dx = integral of [m.dv.v] replace dx = Rd theta (angle is equal to arc/radius) Integrate…get the values of v^2…put them in the first equation…youll get the value of N .:D….really easy 😛
sai vishnu September 12, 2011 at 11:23 am
sir, if the process is given, it would be very helpful for the students.
Started in 2002 by two IITians, Quest is the fastest growing coaching brand in India.
Quest initiated “Power Coaching for IITJEE”, a unique system of IITJEE preparation that maximises student performance by synchronizing studies with Class 11/12.
© 2014 Copyright - Quest Tutorials. All Rights Reserved. Powered by Erachnida