Question

A mass, m, is held by a spring with a stiffness constant k. The potential energy, p, of the
system is given by: p = 1 /2 kx^2 − mgx where x is the displacement and g is acceleration due to gravity.

The system is in equilibrium if dp/dx = 0. Determine the expression for x for system equilibrium.

Answer 1

Answer:

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Answer 2

Given: A mass, m, is held by a spring with a stiffness constant k. The potential energy, p, of the system is given by: p = (1 /2)kx² − mgx where x is the displacement and g is acceleration due to gravity. The system is in equilibrium if dp/dx = 0.

To find: The expression for x for system equilibrium.

Solution:

dp/dx is the differentiation of the equation given by p with respect to x. When we differentiate the (1/2)kx² part, (1/2)k stays as it is since they are constants and x² changes to 2x. When we differentiate the -mgx part, -mg stays as it is since they are constants and x changes to 1. This can be represented as follows.

$p = \frac{1}{2}kx^{2} - mgx$

$\frac{dp}{dt} = \frac{d}{dt} (\frac{1}{2}kx^{2} - mgx)$

    $= \frac{1}{2}k(2x) - mg(1)$

    $= kx - mg$

But the system is in equilibrium if dp/dx = 0.

$kx - mg = 0$

$x = \frac{mg}{k}$

Therefore, the expression for x for system equilibrium is mg/k.