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Carnegie Mellon University

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Problem 21 Medium Difficulty

A horizontal spring attached to a wall has a force constant of $k=8.50 \times 10^{2} \mathrm{N} / \mathrm{m}$ . A block of mass $m=1.00 \mathrm{kg}$ is attached to the spring and rests on a frictionless, horizontal surface as in Figure $\mathrm{P} 13.21 .$ (a) The block is pulled to a position $x_{i}=6.00 \mathrm{cm}$ from equilibrium and released. Find the potential energy stored in the spring when the block is 6.00 $\mathrm{cm}$ from equilibrium. (b) Find the speed of the block as it passes through the equilibrium position. (c) What is the speed of the block when it is at a position $x_{i} / 2=3.00 \mathrm{cm}$ ?

Answer

a. 1.53$\mathrm { J }$
b. 1.75$\mathrm { m } / \mathrm { s }$
c. 1.52$\mathrm { m } / \mathrm { s }$

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Video Transcript

four party. The potential energy of a spring initially would be equal in 1/2 times K X initial squared. It is equal in 1/2 multiplied by 850 newtons per meter multiplied by 6.0 times 10 to the negative second meters quantity squared and this is giving us 1.53 jewels. So I'll be your answer for part a four part B. Then we know the surfaces frictionless so we can apply the law of conservation of mechanical energy. And we can say that then the potential energy of the spring initially will be equaling to the kinetic energy final. And so we can say that 1/2 and the max squared will be equaling 1.53 jewels, so our maximum velocity would be equal in the square root of two multiplied by 1.53 jewels. This would be divided by the mass of 1.0 kilograms and this is giving us 1.75 meters per 2nd 4 hour maximum velocity for part B and for part C. Then here we have an exposition which is equaling half of the initial exposition, so this is equaling six divided by two so three centimeters and so the velocity. At this point, we can say that here, the law of conservation of mechanical energy gives us that the kinetic energy at this particular exposition, plus the potential energy at this particular exposition, is equaling the total amount of energy. And so we know the total amount of energy equaling one point 53 jewels. What we found in part A. And so we can then say that the velocity would be equaling the square root of two times the amount to time the total amount of energy minus K X squared, and this would be divided by the mass M. And so the velocity would be equal in the square root of two multiplied by 1.53 jewels, minus 850 newtons per meter multiplied by 3.0 times 10 to the negative second meters quantity squared. Let's extend the square root and all of this will be divided by the mass of one kilogram. And so the velocity is found to be 1.51 meters per second and this is again given that we have an exposition of three centimetres. So half of our maximum of six centimeters, 1.51 meters per second would be our answer for part C. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
Top Physics 102 Electricity and Magnetism Educators
Marshall S.

University of Washington

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University of Sheffield

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