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When working with a conservative force, it is generally easier to work with a potential energy function, which is a scalar, than to compute the work along some path from the forces and displacements, which are vectors. the change in the potential energy of an object depends only upon its initial and final position, but it is independent of the path undertaken to move between those positions. some effort is initially required to obtain the potential energy function. consider a block of mass m attached to the end of a horizontal hooke's law spring with spring constant k. the block is free to move along a frictionless horizontal surface, and displacements are measured from the equilibrium position of the spring, taken here as the origin of the coordinate system.

a = fs = - k x iì‚
b = ds = dx iì‚
c = dw = - k x dx
d = dus = k x dx

required:
a. a finite change in the potential energy may be obtained by integrating infinitesimal changes. what integral expressions is correct for this hooke's law spring?
b. perform the integrations, and input an expression for the change in the potential energy of the spring, also known as the elastic potential energy.

Sagot :

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a. The integral expression for the change in potential energy of the spring is ΔU = ∫kxdx

b. The change in potential energy of the spring is ΔU = k(x₂² - x₁²)/2

Since we are given the differential elastic potential energy, du, to find the integral expression, we integrate du.

a. The integral expression for the change in potential energy of the spring

The integral expression for the change in potential energy of the spring is ΔU = ∫kxdx

Since the infinitesimal elastic potential energy of the spring du = kxdx, integrating both sides of the expression, we have

ΔU = ∫du

ΔU = ∫kxdx

So, the integral expression for the change in potential energy of the spring is ΔU = ∫kxdx

b. The elastic potential energy of the spring

The change in potential energy of the spring is ΔU = k(x₂² - x₁²)/2

Performing the integration for the change in potential energy of the spring

ΔU, we have that integrating for change in length from x₁ to x₂

ΔU = ∫kxdx

ΔU = k[x²/2]₁²

ΔU = kx₂²/2 - kx₁²/2

ΔU = k(x₂² - x₁²)/2

So, the change in potential energy of the spring is ΔU = k(x₂² - x₁²)/2

Learn more about elastic potential energy here:

https://brainly.com/question/25996974

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