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$\bullet$ $\bullet$ A spring is 17.0 $\mathrm{cm}$ long when it is lying on a table. Oneend is then attached to a hook and the other end is pulled by aforce that increases to 25.0 $\mathrm{N}$ , causing the spring to stretch to alength of 19.2 $\mathrm{cm} .$ (a) What is the force constant of this spring?(b) How much work was required to stretch the spring from17.0 $\mathrm{cm}$ to 19.2 $\mathrm{cm}$ (c) How long will the spring be if the25 $\mathrm{N}$ force is replaced by a 50 $\mathrm{N}$ force?

a) 1136 $\mathrm{N} / \mathrm{m}$b) 0.275 Jc) 0.214 $\mathrm{m}$

Physics 101 Mechanics

Chapter 7

Work and Energ

Physics Basics

Applying Newton's Laws

Kinetic Energy

Potential Energy

Energy Conservation

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

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Okay, so we have a spring with length 17 centimeter and then a Force 25 Newton is applied to the spring, and then it stretched to 19.2 centimeter. So we need to find out the spring constant from this ah force. Now, we know that force on spring is equal to the spring, constant times the length or the stretched length. So in here, X will be the extra length that this being has stressed due to the force. In other words, X will be 19.2 centimetre minus our ending 0.0 centimeter, which is, um, two plain two centimeter are 0.2 meter. Now, if we're gonna figure out the spring constant, that will be force on the Applied Force divided by the stretch. And we know that they applied force is 25 Newton. So we can write that down, and we divide that by the stretch, which is 0.2 meter. From there, we get this spring constant as 1.1 four times and to the bar three newton meters now in part B. Um, were we already know that X is zero points. You're two meter and we need to find out the work done on the spring due to the force so worked. And on spring will be half que existe square again access the stretch and we already know the spin constant from the previous part so we can write down the numbers again. So we'll have 1.14 times 1.14 10 to 3 Newton Meter cube, middle on Newton per meter times 0.2 meters squared. From there we get the work done as 0.276 dues and finally in part. See, ah, we're told that the force is double so if we double the force, we know that forces a relation as a proportion in relation with the stretch. Because Casey Constance, if we double the force that stretch with will be double as well. So initially, the stretch was 0.22 and finally the stretch will be twice off the initial, which is two times zero point 0 to 2 meter, which is 0.0 44 meter and therefore the final velocity with our Sorry, The final force will be 17 centimetre plus 4.4 centimeter. It is 21.4 centimeter. Thank you

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