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Question

# Calculate the work required to stretch the following springs 1.25 m from their equilibrium positions. Assume Hooke's law is obeyed. A spring that requires a force of $250 \mathrm{~N}$ to be stretched $0.5 \mathrm{~m}$ from its equilibrium position.

Solution

Verified

b.

A force of $250$N is required to keep the spring stretched at $x = 0.5$m means by Hooke's law:

$F(0.5) = k(0.5 \text{ m}) = 250 \text{ N}$

Solving for $k$ we get:

$k = 500 \text{ N/m}$

Hence, Hooke's law for this spring is:

$F(x) = 500x$

The spring is compressed from $x = 0$ to $x = 0.4$, so the work is:

$W= \int_0^{1.25} 500x \,dx = \left( 250x^2 \right)\Big \rvert_0^{1.25} = 390.63 \text{ J}$

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