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A block of mass m hangs on a vertical spring. Initially the spring is unstreched, it is now allowed to fall from rest. Find (a) the distance the block falls if the block is released slowely, (b) the maximum distance the block falls before it beigns to move up. |
Answer» <html><body><p></p>Solution :When the block <a href="https://interviewquestions.tuteehub.com/tag/falls-983294" style="font-weight:bold;" target="_blank" title="Click to know more about FALLS">FALLS</a> slowely, it comes to rest at a distance `y_(0)`, which is referred to as the equlibrium <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a>. From, condition of equilibrium, <br/> `sumF_(y)=ky_(0)-<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>=0 ory_(0)=(mg)/(k)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP2_C06_SLV_039_S01.png" width="80%"/> <br/> (b) When the block is released suddenly, it oscilates about the equilibrium position. Initially the speed of the block <a href="https://interviewquestions.tuteehub.com/tag/increases-1040626" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASES">INCREASES</a> then reaches maximum value and then <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> to zero at the lowest position. In this situation the block oscillates about the equilibrium position. The block is released from rest, there fore its total mechanical energy initially. <br/> `E_(i)=U_(g)+U_(s)+K=0+0+0` <br/> Final total mechanical energy, `E_(f)=-mgy_(m)+(1)/(2)ky^(2)+0` <br/> From conservation of energy, `E_(i) = E_(f)` <br/> `0=-mgy_(m)+(1)/(2)ky^(2)ory_(m)=(2mg)/(k)=2y_(0)`</body></html> | |