# Time as a Function of Compression?

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Question-
A block of mass $m$ is kept on a rough horizontal surface (coefficient of friction - $\mu$) and is connected to the wall by a spring of spring constant $K$. A force $F$ is applied to it in the direction of the wall. Find the $time{\kern 1pt} taken$ to compress it by $x$ units.

My Attempt-
When the spring has been compressed by $x$ units, the instantaneous velocity $v$ of the object can be found out through Law of Conservation of Energy, $$Fx - \mu mgx = {\textstyle{1 \over 2}}K{x^2} + {\textstyle{1 \over 2}}m{v^2}$$ $$v(x) = \sqrt {{{2Fx - 2\mu mgx - K{x^2}} \over m}}$$Since $v(t) = {{dx} \over {dt}}$,   so  $dx = v(t)dt$.
I am not able to integrate this equation as my expression for $v$ is a function of $x$.
How do we proceed from here?

related to an answer for: Trouble With Springs-1
asked Mar 29, 2017

## 1 Answer

4 votes

Best answer

$$v(x)=dx/dt = \sqrt {{{2Fx - 2\mu mgx - K{x^2}} \over m}}$$
$$\frac{dx}{\sqrt {2Fx - 2\mu mgx - K{x^2}}} = \frac{dt}{\sqrt{m}}$$
Now integrate it.
By using $\int \frac{1}{\sqrt{a-x^2}}=\arcsin x/a$

answered Mar 29, 2017 by (4,286 points)
selected Mar 29, 2017