Kinetic Energy ```Name: N/A Status: N/A Age: N/A Location: N/A Country: N/A Date: 1991 ``` Question: I am a senior at Monta Vista High School in Cupertino, California and I am currenlty enorlled in Physics. Right now we are dealing with kinectic energy. My question is: You were given a 2Kg mass and you applied a 9 Newtons of force across a frictionless table for 4 sec. starting at rest. What kind of kinetic energy does it have now? Umm.. Think about it and let me know as soon possible. I understand the equations in Physics, but the problem solving is most of the time a problem for me. Replies: from F = m * a we have a = 4.5 m/s^2 then v = a * t = 18 m/s then E = 1/2 * m * v^2 = 324 J John Hawley Kinetic energy = (1/2)*M*v^2 where M = mass, v = speed We know M, but not v, so we need to find v at time t = 4 sec. speed : v = v0 + A*t where v0 is the initial speed (it is 0 here), A is the (constant) acceleration, A = Force/Mass, t = time So, at t = 4 sec., v = 0 + (9/2)*4 = 18 m/sec Thus, K.E. = .5 * 2 * 18^2 = 324 joules rcwinther So, how does one solve such problems in general? Start from what you need to know and work backwards a bit, and start from what you already know, and work forwards until you can get the two sides to meet somewhere: Here - you needed kinetic energy. Did you look up the formula for kinetic energy? That is the first thing you need to do - find out what it is you need to know in order to know what the K.E. is. The formula tells you what you need: mass and speed. Do you already know either of those? Yes! You know what the mass is. But you still need speed. Is there a formula for speed around somewhere? Well maybe it is a good time to start at the other end - you are given a force, and what does force tell you? There is Newton's law F = ma which you should realize by now is the key to almost all force problems. That tells you what a is (since you know the mass m). And so you know a, you need v. Can you find a formula that links the two? You also know initial speed (0) and the time that the force was applied (4 seconds). Well, the others have already given complete answers... Arthur Smith Click here to return to the Physics Archives

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