The potential energy function for a non-linear spring may be written as U(x) = 8

Question

The potential energy function for a non-linear spring may be written as U(x) = 8x- 3x^2 + 1/10 x^3 . The function is in units of Joules when x is measured in centimeters. The spring may be stretched to a maximum length of 2.95 cm. At what position(s) is the spring in equilibrium? Which of these (if any) are stable equilibrium positions? If a 325 g mass is attached to the spring and moves from x = 1.5 cm to x = 2.3 cm, does the spring do the work or must some external force be applied to cause this displacement?

Explanation / Answer

a)
U(x) = 8x -3x^2 +x^3/10
Spring is in equilibriu if dU/dx = 0
dU(x)/dx = 8 - 6x + 3x^2/10
8 - 6x + 3x^2/10 = 0
3x^2 -60x + 80 = 0
solving the above quadratic equation,
x = 18.6 and x =1.4
So at x = 18.6 cm and x =1.4 cm, it is in equilibrium
b)
one of the x correcpond to minimum energy and other to maximum energy
U(x) = 8x -3x^2 +x^3/10
at x = 18.6 cm
U(18.6) = 8*18.6 -3*(18.6)^2 +(18.6)^3/10
= -246 J
at x = 1.4 cm
U(1.4) = 8*1.4 -3*(1.4)^2 +(1.4)^3/10
= -5.6 J
Stable equilibrium are both since energy is negative.
It is more stable at x = 18.6 cm

c)
U(x) = 8x -3x^2 +x^3/10
at x = 1.5 cm
U(1.5) = 8*1.5 -3*(1.5)^2 +(1.5)^3/10
= 5.59 J
at x = 2.3 cm
U(2.3) = 8*2.3 -3*(2.3)^2 +(2.3)^3/10
= 3.75 J
Chnage in energy as it move = 3.75 - 5.59 = -1.84 J
Since change in energy is negative, spring does the work

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