## Engineering Mechanics: Statics & Dynamics (14th Edition)

$9.29ft$
We can determine the required distance as follows: According to the principle of work and energy $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ $\implies \frac{1}{2}(\frac{4000}{32.2})(55)^2+U_{1\rightarrow 2}=0$ This simplifies to: $\Sigma U_{1\rightarrow 2}=187.89kip\cdot ft$ We know that the work done can also be calculated as the area under the curve in the given figure: $\implies \Sigma U_{1\rightarrow 2}=(2\times 9)+(5-2)\times 18+(x\times 27)$ $\implies 187.89=(2\times 9)+(5-2)\times 18+(x\times 27)$ This simplifies to: $x=4.29ft$ Thus, $s=x+5=4.29+5=9.29ft$