## Fundamentals of Physics Extended (10th Edition)

$6.3\dfrac {m}{s^{2}}$
Lets assume that $H_{0}$ is initial height of the box relative to the ground. Total energy of the spring and box system initially will be : $E_{tot}=mgH_{0}..........(1)$ When the box starts to move, its potential energy will be converted to the box’s kinetic energy and spring’s potential energy. Lets assume that box will descend distance $L$ on the incline (so the spring will stretch by distance $L$). Now, lets write total energy equation again: $E_{tot}=E_{spring}+E_{box}=\dfrac {kL^{2}}{2}+\dfrac {mv^{2}_{1}}{2}+mgH_{1}.........(2)$ Here, $H_{1}$ is the height of the box relative to the ground after descending $L$ distance. From the incline, we get: $\Delta H=H_{0}-H_{1}=L\sin \theta \left.......( 3\right)$ From (1) and (2), we get: $mg\left( H_{0}-H_{1}\right) -\dfrac {kL^{2}}{2}=\dfrac {mv^{2}_{1}}{2}\Rightarrow v_{1}=\sqrt {2g\left( Ho-H_{1}\right) -\dfrac {k}{m}L^{2}}\left...( 4\right)$ From (3) and (4) we get: $v_{1}=\sqrt {L\left( 2g\sin \theta -\dfrac {kL}{m}\right) }....(5)$ So if the block momentarily stops, then $v_{1}=0\left.....( 6\right)$ From (5) and (6), we get: $L=\dfrac {2mg\sin \theta }{k}\approx 0.21m...........(7)$ So at that point, lets find the acceleration of the box: $ma=kL-mg\sin \theta$ Using (7), we get $a=g\sin \theta \approx 6.3\dfrac {m}{s^{2}}$