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An "extreme" pogo stick utilizes a spring whose uncom-pressed length is 46 $\mathrm{cm}$ and whose force constant is$1.4 \times 10^{4} \mathrm{N} / \mathrm{m} . \mathrm{A} 60-\mathrm{kg}$ enthusiast is jumping on the pogo stick, compressing the spring to a length of only 5.0 $\mathrm{cm}$ atthe bottom of her jump. Calculate (a) the net upward forceon her at the moment the spring reaches its greatest com-pression and (b) her upward acceleration, in $\mathrm{m} / \mathrm{s}^{2}$ and $g^{\prime}$ s atthat moment.

a) 5100 $\mathrm{N}$b) 85 $\mathrm{m/s^2}$ and 8.70$\mathrm{g}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

Hope College

McMaster University

Lectures

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2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

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Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

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question 57 and extreme Pogo Stick utilizes a spring whose UN compressed length is 46 centimeters and whose horse constant is 1.4 times 10 to the four Newtons per meter. A 60 gram, 60 kilograms enthusiast is jumping on the pogo stick person in the spring to a length of only five centimeters at the bottom of her jump a Catholic, the net upward force on her at the moment this straight spring reaches its greatest compression and be her upper upward acceleration and meters per second and G's at that moment. All right, so we have the scenario where the spring is being compressed. Um, when she's pushing her weight all way down this, of course, they'll be a force acting on this spring to pull it back up. So the net the net force due to the weight pushing down in there, be a restoring force. F Spring will say acting upward. So for part may for you to find this the Net force Was she acting on upward on the spring would be spring force itself and minus the weight based on our free our representation of free body diagram here at the bottom, right? So for our, um, we took into accounts our restoring force. First, the direction of our spring concert doesn't such that it's positive que x my weight. Which is f g, of course, the amount of spring presses X So the compression of the spring simply its initial X initial minus X final. 46 centimeters, minus five centimeters. I just Did they just be 41 senator years and for this millimeters 7.41 meters So we have all our term now are for its constant. One point for that has tended before tunes per meter Multiply it by compression again Point for what meters? Attracting the mass of the, uh posted a user If we assume that this is state question but we can assume that the mass of the pogo stick is negligible here Times 9.8 Chief 9.8 Major second, which is this one g. We find to two significant figures running down that are net force. Acting on the holistic is 5100 Newtons, or B if you want to find the acceleration well, we know the net force acting on a pogo stick It simply I m A which I just wrote is a M for some reason I mean because I want to divide both sides by em so that the acceleration is the force took over the mass. So in meters per second, we can calculate this Teoh 5100 Newtons have every 60 kilograms. Would you give the value of their exploration to two significant figures to 85 meters per second squared. Additionally, if you want to find, uh, exploration G well, we can also straight in terms of weight over G. Well, actually, no, it slightly easier what we can do if you just take our value for acceleration. Divide that by G then 85 side by 9.8. And so our final value in terms of G R acceleration is 8.70 Our guest is 8.7 to 600 figures here, 8.7 times g. So these two values are corpulent. There you go

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