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Question

A 6-kg box on a frictionless horizontal surface is attached to a horizontal spring with a force constant of $800 \mathrm{~N} / \mathrm{m}$. If the spring is stretched $4 \mathrm{~cm}$ from its equilibrium length, what is the acceleration of the box?

Solution

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Answered 1 year ago

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We have:

- A box with mass $m=\mathrm{6\hspace{1mm}kg}$;
- A spring with constant $k=\mathrm{800\hspace{1mm}\frac{N}{m}}$;
- Spring's stretching $\Delta l=\mathrm{4\hspace{1mm}cm=4 \cdot 10^{-2}\hspace{1mm}m}$.

We need:

- Acceleration of the box $a$

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