Question

A clever engineer designs a "sprong" that obeys the force law fx =−q(x−xe)3, where xe is the equilibrium position of the end of the sprong and q is the sprong constant. for simplicity, we'll let xe =0m. then fx =−qx3.

Answer 1

The objective of this question is to compute the units of q and calculate the expression for the P.E of the compressed spring

Answer:

q = N/m³

$\mathbf{U(x) = -\dfrac{qx^4}{4}}$

Explanation:

Given that:

The equation for the force law is:

$F_x = -q (x -x_e)^3$

where;

$x_e$e = equilibrium position of the end of the spring

i.e. $x_e$ = 0

q = is the spring constant

Also;

$F_x = -qx^3$

From above the units of q can be calculated by making q the subject of the formula :

$q = \dfrac{F_x}{x^3}$

where x is in meters and F is in Netwon;

Then :

q = N/m³

The Potential energy P.E of the compressed spring can be calculated by using the integral:

$F_x = -\dfrac{dU}{dx}$

$U(x) = - \int F_x dx$

$U(x) = - \int ^x_0 (-qx^3) \ dx$

$\mathbf{U(x) = -\dfrac{qx^4}{4}}$

Answer 2

The unit of q is N/m³.While the potential energy of the compressed spring will be $\rm U(x)= -\frac{qx^4}{4}$.

What is the potential energy of spring?

When a spring deviates from its mean position, it attempts to regain equilibrium by applying a force that is equivalent to but opposite to the external force.

Spring force has been used in bicycle carriers and launching mechanisms, where the energy generated by disrupting the spring's balance is used as potential energy.

The given equation for the force law is;

$\rm f_x =-q(x-x_e)$

Where,

$\rm x_e$ is equilibrium position of the end of the spring

q is a spring constant

If at equilibrium position;

$\rm x_e=0$

$\rm F_X= -qx^3 \\\\ \rm q = \frac{F_x}{x^3}$

So the unit derived will be;

q = N/m³

The potential energy of the spring is found by;

$\rm F_x=- \frac{du}{dx} \\\\ U(x)= -\int\limits^x_0 {F(x)} \, dx \\\\ U(x)= -\int\limits^x_0 -qx^3 \, dx \\\\ U(x)=- \frac{qx^4}{4}$

Hence the unit of q is N/m³.While the potential energy of the compressed spring will be $\rm U(x)= -\frac{qx^4}{4}$.

To learn more about the potential energy of spring refer to the link;

https://brainly.com/question/2730954