question_answer

a block of mass m is pushed against a spring of spring constant k fixed at one end to a watt. The block can slide on a frictionless table as shown in Eig.4(HT).8. The natural length of the spring is lq and it is compressed to half its natural length when the block is released. Find the velocity of the block as a function of its distance x from the wall.   

Answer:

                On releasing the block, the spring pushes the block tothe right till the spring acquires its natural length\[\left( {{L}_{0}} \right)\]. At this stage, the block loses contact with the spring and moves with a constant velocity. Initial compression of spring = \[\frac{{{L}_{0}}}{2}\]. When block is at a distance x from the wall, where \[x<{{L}_{0}}\], the compression is\[({{L}_{0}}-x)\]. Using the principle of conservation of energy. \[\frac{1}{2}k{{\left( \frac{{{L}_{0}}}{2} \right)}^{2}}=\frac{1}{2}k{{\left( {{L}_{0}}-x \right)}^{2}}+\frac{1}{2}m{{\upsilon }^{2}}\] \[\upsilon =\sqrt{\frac{k}{m}}{{\left[ \frac{L_{0}^{2}}{4}-{{\left( {{L}_{0}}-x \right)}^{2}} \right]}^{1/2}}\]