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$\bullet$ An unstretched spring is 12.00 $\mathrm{cm}$ long. When you hang an875 g weight from it, it stretches to a length of 14.40 $\mathrm{cm}$ .(a) What is the force constant (in $\mathrm{N} / \mathrm{m}$ ) of this spring?(b) What total mass must you hang from the spring to stretch itto a total length of 17.72 $\mathrm{cm} ?$

a) 357 $\mathrm{N} / \mathrm{m}$b) 2.08 $\mathrm{kg}$

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

University of Sheffield

University of Winnipeg

Lectures

04:01

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.

03:28

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. 55 states that unstrapped spring is 12 centimeters long. You hang 835 grams from it, it stretches to length of 14.4 centimeters. Part of this question asks, What is the force constant of this spring? And B? What total mask must you hang from the spring to stretch it to a total length of 17.7 Do centimeters? All right, so this scenario where I would say it may be the spring is attached to the ceiling. Is there spring here in the attached a mass to it. So this mass and of course, a pool the object down due to its weight. However, because spring is restoring force, the spring will want to pull itself back towards normal on Stretch State. Some call this force of the spring so these two forces counter act each other and balance such that there's no more movement in the spring once the oh X once the spring is attached. So for your first letter favorable axes, we know that in this scenario, the net force acting on the system has to be zero. Because this system is the axillary. That means we have the force of the spring for the upward and the weight going down, which has equal because zero it's bring for us, which we take the direction with. A sign of it into account in the problem because we're pointing upward was pulling its up away from the stretching scenario it is. You make that call them to K X by Hook's Law, which equals the attached weight of the object with MGI to for myself for the, um, force constant in the spring. We can just re registration to BMG over X, but we're not given directly how much the stream stretches were just given an initial state in the final state of the spring length. So if we plug in our values now, so a massive 0.875 I mean, this should be kilograms, of course times 9.8 meters per second squared for value for gene on our distances. Our final stretch is 14.4 centimeters minus 12 centimeters. We did it with advisers by 100 to convert that two meters, of course, so, really, the spring distance distress distances just 14.4 minus 12. So the two point for CME, which is a value for the denominator, however, with the comfort that two meters, 0.0 24 meters which should be a value on your nominee near. So once you do that, we can calculate our spring constant are forced constant to be 357 newtons per meter. Great Herbie says that now the total we're pulling this spring are X final now 17.72 centimeters which now that means if we start from the initial position of 12 centimeters are total stretch distance just given by ex A 17.7 to minus 12 should just be 5.72 centimeters and equitably 0.5 72 meters. And so, by the same logic from part any of this question this spring constant or which we know Sorry, that won't change this time. We're looking to solve for a mass. So rearranging this equation the mass is gonna by K X over g. So in the spring concert, the force constant sorry is constant for the string spring, the better. How much of stretches? Okay, I made the same. The distance is zero point zero. If I seven to meters that all you can about the value found here. Let me divide that by are known life for G 9.8 years per second. Thank you meters per second squared. And so by calculating this, we can find that the necessary master attached to this the three significant years is too went 08 kilograms. You

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