using the work-kinetic energy theorem:

.5mvo^2 + Fo/c = .5 mv1^2 where v1 is the velocity at x max I got the following but it is not formatted proprerly. Where am I wrong?

.5vo^2 +Fo/c*m = .5v1^2
vo^2 + 2Fo/c*m = v1^2
sqrt (vo^2 + 2Fo/c*m) = v1

4 answers

What is c?
m to the -1 power
c= m to the -1 power
then I don't understand your second term:

Fo/c*m if c= m^-1, then
Fo/(1/m *m)=Fo

which is not right. YOu cant add Fo to KE to have work energy.
Similar Questions
  1. Kinetic Energy and the Work/Energy Theorem:1.) How much kinetic energy does a 700-gram baseball have that is travelling at 20
    1. answers icon 5 answers
    1. answers icon 1 answer
    1. answers icon 2 answers
  2. true or false1. The work required to stretch a spring changes linearly with the amount of strech? 2. The kinetic energy of an
    1. answers icon 1 answer
more similar questions