Anonlinear spring model could include a cubic term. that is, the force is approximated by f(x) = –kx + k’x3 . a) if all quantities are in our usual (si) units, what are the units of k and k’? b) for a particular spring, the constants are k = 300 and k’ = 5x104 . how much does the potential energy in the spring change when it is stretched from x = 0.02 m to 0.05 m?

3548.75 x 10⁻⁴  J

Explanation:

F(X) = - kx + k⁻x³

unit of kx = unit of F ( force )

kx = Newton (N)

k m = N

unit of k = Nm⁻¹

Similarly

unit of

k’x3 = N

k’m³= N

k’ = Nm⁻³

Work done in stretching spring by x will be equal to potential energy

F(x) = - 300 x + 5 x 10⁴ x³

work done

= ∫ F(x) dx

= ∫( - 300 x + 5 x 10⁴ x³ )  dx

=

= - 300 x ( .02 )² / 2  + 5 x 10⁴ x ( .02 )⁴/4 - ( - 300 x ( .05 )² / 2  + 5 x 10⁴ x ( .05 )⁴/4 )

=  -3548.75 x 10⁻⁴

potential energy stored = 3548.75 x 10⁻⁴  J

the answer would be a

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.