Compression of spring when an object of given mass is placed on it - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-23T07:19:40Z https://physics.stackexchange.com/feeds/question/209884 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/209884 0 Compression of spring when an object of given mass is placed on it MayankJain https://physics.stackexchange.com/users/81999 2015-09-30T05:24:48Z 2016-07-05T19:22:37Z <p>A question in my textbook says - A block of mass $m$ is released from rest onto an ideal non deformed spring from a negligible height. Neglecting air resistance, find compression $x$ of the spring. I tried using two approaches for this Q but I can't figure out which one if right or if both are wrong. </p> <p>1st - </p> <p>Using Hooke's Law. $$mg -kx = 0$$ $$x = (mg) / k$$</p> <p>2nd - Using Work-Energy theorem. Considering the Gravitational potential energy at the lowest point of the compression to be 0 and since change in KE is 0. </p> <p>$$mgx -(kx^2)/2 = 0$$ $$mgx = (kx^2)/2$$ $$x = (2mg)/k$$</p> <p>Is any of my approach correct? If yes then why is the other one wrong? Or are both completely wrong? </p> https://physics.stackexchange.com/questions/209884/-/209890#209890 1 Answer by sarat.kant for Compression of spring when an object of given mass is placed on it sarat.kant https://physics.stackexchange.com/users/91917 2015-09-30T06:09:38Z 2015-09-30T06:09:38Z <p>Both are right. The first approach gives the compression where the <strong>net force</strong> on the object is <strong>zero</strong>. The second approach gives the compression when the <strong>velocity</strong> of the object is <strong>zero</strong>. When the block falls on the spring, it oscillates between $x=\frac{2mg}{k}$ and $x = 0$. Since the spring is ideal and the air resistance is negligible, this oscillation does not die down and so the question is wrong. But if the oscillation dies down eventually, the spring comes to rest at $x = \frac{mg}{k}$.</p> https://physics.stackexchange.com/questions/209884/-/209893#209893 1 Answer by John Rennie for Compression of spring when an object of given mass is placed on it John Rennie https://physics.stackexchange.com/users/1325 2015-09-30T06:29:54Z 2015-09-30T07:44:23Z <p>The work-energy theorem is certainly the easiest way to do the problem, but you can also solve it by calculating the force.</p> <p>In any situation where you need to calculate the response of an object to a force you use Newton's second law. This tells us (after a minor rearrangement):</p> <p>$$\frac{d^2x}{dt^2} = \frac{F}{m} \tag{1}$$</p> <p>In this case the force on the mass has two parts. There is a downwards force of $F_{gravity} = mg$ due to gravity, and an upwards force due to the spring. Assuming the spring obeys Hooke's law then the force when the spring is compressed a distance $x$ is given by:</p> <p>$$F_{spring} = -kx$$</p> <p>The net force on the mass if $F_{gravity} + F_{spring}$, and putting this into equation (1) we get:</p> <p>$$\frac{d^2x}{dt^2} = \frac{mg - kx}{m}$$</p> <p>If you solve this differential equation you get an equation for $x$ as a function of time. It's then straightforward to find the maximum value of $x$.</p> <p>The equation is a lot easier to solve than it looks, because it's the equation for a simple harmonic oscillator in disguise. However if you haven't done simple harmonic motion yet this is probably not a route you want to pursue.</p> https://physics.stackexchange.com/questions/209884/-/266405#266405 0 Answer by samarth srivastava for Compression of spring when an object of given mass is placed on it samarth srivastava https://physics.stackexchange.com/users/122800 2016-07-05T19:22:37Z 2016-07-05T19:22:37Z <p>according to me you have erred in assuming that change in KE is 0 as the block will have some velocity when it is compressed.since total mechanical energy remains constant loss in P.E will result in gain of K.E</p>