How Much Work is Required to Stretch a Spring 3 Meters from Equilibrium?

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SUMMARY

The discussion focuses on calculating the work done in stretching a spring 3 meters from its equilibrium position using Hooke's Law. Given that a spring can be stretched 5 meters with a force of 50 N, the spring constant k is determined to be 10 N/m. The work done (W) is calculated using the formula W = (1/2)kx^2, resulting in 45 Joules for the 3-meter stretch. Additionally, the discussion includes evaluating the improper integral $\int_{0}^{\infty} \frac{dx}{x^2-16}$ using partial fraction decomposition and knowledge of inverse hyperbolic functions.

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my final is tomorrow and my instructor gave a list of questions that will be similar to the ones on the final exam and i want to see how they should be done properly. I've been working on other problems but i can't get past these ones. thanks
a spring on a horizontal surface can be stretched 5m from equilibrium position with force 50 N. how much work is done stretching the spring 3m from the equilibrium position?evaluate the integral
$\int_{0}^{\infty} \ \frac{dx}{x^2-16}$
 
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ineedhelpnow said:
a spring on a horizontal surface can be stretched 5m from equilibrium position with force 50 N. how much work is done stretching the spring 3m from the equilibrium position?
Hooke's Law says F = kx (using magnitudes). So you can use the initial information to find k. Then W = (1/2)kx^2.

ineedhelpnow said:
evaluate the integral
$\int_{0}^{\infty} \ \frac{dx}{x^2-16}$

Hint: Use a partial fraction decomposition:
\frac{1}{x^2 - 16} = \frac{A}{x + 4} + \frac{B}{x - 4}.

-Dan
 
For the improper integral, you can also factor -1 out of the denominator which will turn it into a very familiar antiderivative :)

EDIT: It's only familiar if you know the derivatives of inverse hyperbolic functions.
 

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