Physics
posted by Alex on .
A thrillseeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies
Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its highest point.

Solve:
vbr{lr<5vvx><f(nn)/.bbn>
vbr<lr><5.12>(9.8)^3
.bbn=%ff
%ff=F(9x)
.bbn=%ff9x
.bbn=.133%tri
.bbn=.tri
t.ri=vbr
v=31.4
b=59.3
r=28.6
KE___? 
^^^Wrong!
10c<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C<0.09>=f(1.33)
ANS:___? 
dude I have no idea what you just typed..
can you please clarify what all the variables you used stand for? 
fl= flux
uf= kinetic flux
fx= potential flux
l= lib
C= constant