Chapter 7 Problem 57

Problem

A mass $m$ is attached to a spring which is held stretched a distance $x$ by a force $F$, and then released. The spring compresses, pulling the mass. Assuming there is no friction, determine the speed of the mass $m$ when the spring returns:

(a) to its normal length ($x=0$);

(b) to half its original extension ($x/2$).

Solution

(a)

The spring constant is found by the magnitudes of the initial force and displacement, and so

$k=F/x$

As the spring compresses,it will do the same amount of work on the block as was done on the spring to stretch it. The work done is positive because the force of the spring is parallel to the displacement of the block. Use the work-energy theorem to determine the speed of the block.

$W_{\begin{array}{l}{\text{on block}}\\{\text{compression}}\end{array}}=\Delta K_{\text{block}}=W_{\begin{array}{l}{\text{on spring}}\\{\text{stretch}}\end{array}}$ $\rightarrow {\frac {1}{2}}mv_{f}^{2}={\frac {1}{2}}kx^{2}={\frac {1}{2}}{\frac {F}{x}}x^{2}$ $\rightarrow v_{f}={\sqrt {\frac {Fx}{m}}}$

(b)

Now we must find how much work was done on the spring to stretch it from $x/2$ to $x$. This will be the work done on the block as the spring pulls it back from $x$ to $x/2$

$W_{\begin{array}{l}{\text{on spring}}\\{\text{stretch}}\end{array}}=\int _{x/2}^{x}Fdx=\int _{x/2}^{x}kxdx=\left.{\frac {1}{2}}kx^{2}\right|_{x/2}^{x}={\frac {1}{2}}kx^{2}-{\frac {1}{2}}k\left(x/2\right)^{2}={\frac {3}{8}}kx^{2}$

${\frac {1}{2}}mv_{f}^{2}={\frac {3}{8}}kx^{2}\rightarrow v_{f}={\sqrt {\frac {3Fx}{4m}}}$