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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 :

Answer:

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

=[tex][  - 300\times \frac{x^2}{2}   + 5 \times 10^4\times \frac {x^4}{4}  ] ^{.02}_{.05}[/tex]

= - 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

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