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1 year ago

Please help with this question, step by stepThank you!

Physics

1 answer:

mamaluj [8]1 year ago

7 0

From the calculation, the force constant is 192 N. Also, friction would decrease the extension.

<h3>What is the force constant?</h3>

We know that the force constant can be obtained by the use of the relation;

F = Ke

F = applied force

K = force constant

e = extension

We know from Hooks law that the force applied is directly proportional to the extension.

We can write;

F = mgcosθ

F = 43 Kg * 9.8 m/s^2 * sin31°

F = 217 N

K = 217 N/1.13 m

K = 192 N/m

If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.

Learn more about Hooke's law:brainly.com/question/14140269

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$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

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First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: