Deriving Distance from the Energy Conservation LawDate: 6 Jul 1995 16:27:59 -0400 From: Alberto Carvalho Peret Subject: Integration Hello, I'm a 1st year materials engineering student here in Brazil, and I have a doubt concearning a physics integration: How do we get the S = So+0.5*v*t^2+Vo*t formula, for a free-falling object, from the energy conservation law? I can start at Torricelli's Equation V^2=Vo^2+2*a*(S-So) I thank you in advance. Caio Peret Date: 12 Jul 1995 13:52:24 -0400 From: Dr. Ethan Subject: Re: Integration Hey Alberto, Here is the answer written by a specialist that I called in on the case. He is a friend of mine in Engineering here at Swarthmore College. He wrote: First the equation governing the kinematics of freely falling objects is usually written y = yo + vo t - 1/2 g t^2 (1) where y is the current displacement as measured from the ground, yo is the starting displacement from the ground, vo is the initial velocity, g is the acceleration due to gravity, and t is the time. The negative sign in the equation indicates that acceleration is downward, so in this equation g = 9.80 m/s^2. Your notation is somewhat different, but since I've clarified each of my variables, you should be able to figure out their equivalents in your textbook. As you mention, an equivalent kinematic equation is v^2 = vo^2 - 2 g (y - yo) (2) where all variables are defined above. You can obtain Equation (1) by substituting the expression for velocity with constant acceleration v = vo - g t (3) into Equation (2) and reducing algebraically. You indicated that you wished a derivation of Equation (1) from energy conservation, but the derivation in the preceding paragraph doesn't really offer an intuitive energy derivation. Instead I would recommend beginning with the mechanical energy conservation law, which is KEo + UEo = KE + UE (4) where KEo is the initial kinetic energy, UEo is the initial potential energy, and KE and UE are the current kinetic and potential energies. Assuming that a particle begins with an initial displacement and velocity and is subjected to gravity, Equation (4) may be written 1/2 m vo^2 + m g yo = 1/2 m v^2 + m g y (5) where m is the mass of the object and all other variables are defined above. In this expression substitute Equation (3) and divide out the mass to give 1/2 vo^2 + g yo = 1/2 (vo - g t)^2 + g y. (6) Solving this equation for y will yield Equation (1), directly from mechanical energy conservation. Hopefully the explanation has been useful to you! ethoen1@cc.swarthmore.edu 256 comin' my way! |
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