Question

A nonlinear 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?

Answer 1

3548.75 x 10⁻⁴ J

Step-by-step 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^2)/(2)   + 5 * 10^4* \frac {x^4}{4}  ] ^(.02)_(.05)`

= - 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