The work done can be calculated asĮxample - Work done when lifting a Brick of mass 2 kg a height of 20 m above ground Example - Constant Force and WorkĪ constant force of 20 N is acting a distance of 30 m. The work is the product force x distance and represented by the area as indicated in the chart. The work done by a constant force is visualized in the chart above. The unit of work in SI units is joule (J) which is defined as the amount of work done when a force of 1 Newton acts for distance of 1 m in the direction of the force. S = distance object is moved in direction of force (m, ft) The amount of work done by a constant force can be expressed asį = constant force acting on object (N, lb f) Don't take that to mean all of this is meaningless though, because understanding simple cases and simple relationships is crucial to understanding complex relationships.When a body is moved as a result of a force being applied to it - work is done. You always have to be very careful what assumptions you're making, because kinetic energy isn't always directly related to an acceleration, it just works out in the one dimensional case. However, if we assume that everything is acting in one line, or is one-dimensional, then everything works as a scalar, and the math actually works out quite nicely without having to use calculus and vector math. Unfortunately, as eczeno mentioned, the math is a little bit tricky because we're relating scalars and vectors. You can convert those to have one term with a change in position over a change in time, over a change in time (acceleration) through algebraic manipulation. However, since you have a change in velocity squared you have two rates with time two terms with "change in position over change in time" (speed) terms. You're correct in saying a difference in velocity does not represent acceleration, because you need to know how fast that change takes place. Hopefully that clears up how you can have acceleration based on change in kinetic energy. Now you've got a change in velocity over a change in time: a rate, which does define acceleration, unlike the (v_f - v_i) term alone (as you said). I'm going to go back to the algebra I linked to, but I'm going to focus more on what you're asking, hopefully making it clearer. The change in kinetic energy does in fact contain acceleration and you do not need the power part to make it into a rate. what i have just said simplifies the issue too much, it's not the whole picture, but it should help bring it into focus a little. thus, acceleration can tell us a lot about the kinetic energy, but if we try to get the acceleration from the change in KE, we cannot retrieve any information about the direction. acceleration, on the other hand, cares about both parts of the velocity. It turns out that kinetic energy is related to the speed of the object, but speed is only part of velocity, there is also the direction, which kinetic energy simply does not care about. The link above is a small step in the direction of mathematical rigor. It is unfortunate that these concepts are introduced in a way that does not always make those connections clear, but that is necessary to reduce the mathematical rigor. these approaches must be connected, because they describe the same physical system, but the connections can be hidden behind some complicated math. Vector dynamics (force, acceleration, velocity), and energy, which is a scalar approach. One way to think about it is that we are approaching the problem from two directions. If not, please tell me so I can rephrase. We have the work done, which is equal to Kinetic Energy, a change in velocity but also, over time, which is acceleration. So, I don't understand how we can use Kinetic Energy to answer a question like this for example: Assuming that all the power of the motor is used to accelerate the cars, calculate how long they will take to reach their maximum speed.įor the question above though, we are given the power, is it correct to say therefore that:Īcceleration = Rate of change in Velocity, a change in velocity, is equal to a change in Kinetic Energy, which is equal to work done. Ok, I understand that a acceleration, a change in velocity, will change the Kinetic Energy since v is part of the Kinetic energy equation.īut acceleration is not simply a change in velocity. So, let's go straight to the point: a = v - u / t
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