The elastic potential energy E is given by E=C p^2 g^2 A l0^3,where p is the density of the metal.g is the acceleration of free fall. A is the cross-sectional area of the wire and C is a constant. determine the si base units of C? Am Confused...help.

Answer 1

E is in kg-m^2/s^2
p is in kg/m^3
A is in m^2
g is in m/s^2

So, we have

kg-m^2/s^2 = kg^2/m^6 m^2/s^4 m^2 C
kg-m^2/s^2 = kg^2/m^2-s^4 C
C = (kg-m^2/s^2)(m^2-s^4/kg^2)
= m^2-s^2/kg

Answer 2

you forgot l=m