(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass m, velocity v, spring constant k, and position x. (Submit a file with a maximum size of 1 MB.)

1/2mv² = ke²

Explanation:

Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;

K.E = 1/2mass×velocity² i.e 1/2mv²

Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²

E.P = 1/2kx²

The equation describing the case where the kinetic energy is twice the elastic potential energy will be;

K.E = 2EP... 1)

Substituting the KE and EP formula into (1), we have;

1/2mv² = 2(1/2ke²)

1/2mv² = ke² which gives the required equation

measure their densities same density, same sub

a.   mass number from atomic number