UCSB Science Line If you increase the speed of a 2.0-kg air puck by 3.0 m/s in 4.0 s, what force do you exert on it? Question Date: 2004-03-03 Answer 1:Well, force is mass times acceleration, and acceleration is the change in velocity over (or divided by) time. So, we get: a = 3.0/4.0 = 0.75 m2/s and F = m*a = 2.0*0.75 = 1.5 N Pick some different values for mass, velocity, and time and try the calculation yourself. Answer 2:The acceleration of the puck is the change in velocity divided by the time interval needed to change the velocity hence acceleration. In your example is: a = 3 m/s/4s= 0.75 m/s2 Then from F = m a, the force is F= (2 kg)*(0.75 m/s2) = 1.5 kg*m/s2 = 1.5 Newtons. Answer 3:This is a simple problem for which you need to use 2 different equations. Force = mass x acceleration (F = M x a) and acceleration = velocity (speed in a particular direction) / time taken(a = V/t). First calculate the acceleration of the puck using the second equation a = V/t so...... a = 3/4 = 0.75 Now use this in the 1st equation to calculate the force F = M x a so... F = 2.0 x 0.75 and the answer is, F = 1.5N Force is measured in Newtons. Best wishes. Answer 4:So, I'm glad you are studying physics! Let's look at what this problem is asking: "If you increase the speed of the puck BY 3.0 m/sec" - ok, so that means that the CHANGE in speed is 3.0 m/sec. It does not matter what the speed was before and after, just that the CHANGE in speed is 3.0 m/sec. We call this change "delta". So "delta-v" is 3 meters/second. That is our first clue. Now we take this clue, and think about the problem - what is it really all about? Well, something with a mass is going at some initial velocity, and something is done to it to make it change its velocity, and that takes place over a certain interval of time. Aha - we say - this is a momentum conservation type of question! The moving mass, in this case an air puck, has an initial momentum, (mass x velocity) before the problem "begins" and then after the problem is over, it has some new momentum (same mass x different velocity). We know, from Newton's second law, that it takes a force to change a body's momentum! And that force has to act over a certain time interval. So... we put this together in math language: F = force dt = time interval m = mass dv = change in velocity F dt = m dv --> F = m dv/dt and that is our final equation (REMEMBER TO ALWAYS SOLVE THE ALGEBRA FIRST BEFORE YOU PLUG IN THE NUMBERS, NO MATTER WHAT YOUR JUNIOR HIGH TEACHER DOES - IT IS BEST TO ONLY INSERT THE ACTUAL NUMBERS IN THE FINAL STEP!!!) NOW we can plug in the numbers: From the given s, m = 2.0 kg, dt = 4.0seconds, dv = 3.0 m/sec F = (2.0 kg)(3.0 m/s) / (4.0 sec) = 1.5 kg-m/sec2 or 1.5 Newtons. Hope this helps. It is important, even in the beginning stages of learning physics, to use GOOD PROBLEM SOLVING strategies! Answer 5:If you increase the speed of a 2.0-kg air puck by 3.0 m/s in 4.0 s, what force do you exert on it? This sounds a lot like a homework question so I'm not going to give the exact answer but, as you know, Newton's second law says Force = Mass X Acceleration. In addition, if you assume constant acceleration, Acceleration = Change in Velocity / Time of Change. You should be able to get the answer from this... Answer 6:This is perhaps the most straightforward physics question that you could ask. According to Newton's second law, force is defined as the mass of an object times its acceleration, which can be expressed mathematically as F = ma. Acceleration is simply defined as the change in velocity over time. The mass that you have given is 2.0 kg. The change in velocity is 3.0 m/s, and the time in which said acceleration took place is 4 s. Therefore, the acceleration is (3.0 m/s)/(4.0 s) = 0.75 m/s2. The force, therefore, would be 2.0 kg * 0.75 m/s2 = 1.5kg*m/s2, or 1.5 Newtons of force (Because Issac Newton defined "force", the standard international unit of force has been named after him. 9.8 Newtons is about 2.2 pounds, 1.5 Newtons is a little bit less than half a pound.Click Here to return to the search form.    Copyright © 2017 The Regents of the University of California, All Rights Reserved. UCSB Terms of Use