User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Mar 30th 2010, 11:10 AM #1 Junior Member   Join Date: Mar 2010 Posts: 2 Force applied to falling object Hey guys how are you all. Lets see, I'm not a physicist (mathematician actually, same name on MHF) but I'm involved in this discussion (some may be aware...) and would like some physics help. (This is not an assessment, tutorial, university, school etc question. Here's the scenario. (Hope I word everything right) An object with a weight of 100N has a force applied to it so that it takes 4 seconds to fall 0.5m at a constant speed. We then repeat except... The same object then has a force applied to it so that it takes 0.5 seconds to fall 0.5m at a constant speed. How do I calculate the forces applied (I guess they would be applied upwards) on the object in both scenarios? To save you time... An object free falling 0.5m takes 0.3192754285 seconds. Thanks guys!   Mar 31st 2010, 12:08 PM #2 Junior Member   Join Date: Mar 2010 Posts: 10 Hello Deadstar. This isn't really my field, I'm just going around on the forum while I wait for an answer to a question of mine, but I think I understand what you want to know. The only thing I don't understand is the "constant speed" thing. If the problem is asking you to obtain constant speed, I guess the solution would be to counteract the gravitational force with the same amount of 100N, so that there would be no force acting on the body, and then push it down to give it a speed of 0.5/4. For the following I'm assuming that you meant constant acceleration, sorry If I'm not getting something. Speed is the derivative of space in respect to time: v = ds/dt, and we also know that in the case of constant acceleration it will be the product of acceleration and time: ds/dt = a * t Acceleration is given by the force acting on the object divided by its mass, so ds/dt = F/m * t We can solve this differential equation with separation of variables and obtain : s = (F/m) * (t^2 / 2) Now we just substitute our data : 0.5 = F/m * 16/2 That gives us : F = m/16 (where m is the mass of our object) This total force we need to act on the object is the sum of the gravitational force and the force we are applying (let's call it F1): m/16 = (9.8*m + F1) F1 = m (1/16 - 9.8) = -9.7375 * m The object has a weight of 100N, so 100N = m*9,8 ----> m = 10.2 Kg F1 = -9.7375 * 10.2 = 99.32 N For the second case you just substitute the different data. I hope this is right, do you have the solution? Last edited by ambie; Mar 31st 2010 at 12:10 PM.   Apr 5th 2010, 12:44 PM #3 Junior Member   Join Date: Mar 2010 Posts: 2 Hey ambie sorry for the late reply. I actually did mean constant speed though. This is all part of an epic discussion on the mechanics of weight lifting. See here for more info. Force applied on a falling mass (not the discussion, that's on another forum)  Tags applied, falling, force, object Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post reventon703 Kinematics and Dynamics 1 Apr 11th 2012 05:47 AM elusive1324 Kinematics and Dynamics 1 Jan 9th 2012 04:53 AM Eva Kinematics and Dynamics 25 Oct 12th 2009 12:09 PM tweedydaf Kinematics and Dynamics 2 Sep 18th 2008 10:33 PM colombo Kinematics and Dynamics 1 Sep 6th 2008 09:41 PM 